NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

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Q. 2.1 
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Ans.
(i) Mass of one electron = 9.10939 × 10–31 kg

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes Number of electrons that weigh 9.10939 × 10–31 kg = 1

Number of electrons that will weigh 1 g = (1 × 10–3 kg)

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 0.1098 × 10–3 + 31

= 0.1098 × 1028

= 1.098 × 1027
(ii) Mass of one electron = 9.10939 × 10–31 kg

Mass of one mole of electron = (6.022 × 1023) × (9.10939 ×10–31 kg)

= 5.48 × 10–7 kg

Charge on one electron = 1.6022 × 10–19 coulomb

Charge on one mole of electron = (1.6022 × 10–19 C) (6.022 × 1023)

= 9.65 × 104 C

Q. 2.2
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10–27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Will the answer change if the temperature and pressure are changed?

Ans.
(i) Number of electrons present in 1 molecule of methane (CH4)

{1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 × 1023 molecules of methane

= 6.022 × 1023 × 10 = 6.022 × 1024
(ii) 
(a) Number of atoms of 14C in 1 mole= 6.023 × 1023

Since 1 atom of 14C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14C is (6.023 × 1023) ×8.
Or, 14 g of 14C contains (6.022 × 1023 × 8) neutrons.

Number of neutrons in 7 mg:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 2.4092 × 1021
(b) Mass of one neutron = 1.67493 × 10–27 kg

Mass of total neutrons in 7 g of 14C

= (2.4092 × 1021) (1.67493 × 10–27 kg)

= 4.0352 × 10–6 kg
(iii) 
(a) 1 mole of NH3 = {1(14) + 3(1)} g of NH3

= 17 g of NH3

= 6.022× 1023 molecules of NH3

Total number of protons present in 1 molecule of NH3

= {1(7) + 3(1)}

= 10

Number of protons in 6.023 × 1023 molecules of NH3

= (6.023 × 1023) (10)

= 6.023 × 1024

⇒ 17 g of NH3 contains (6.023 × 1024) protons.

Number of protons in 34 mg of NH3

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 1.2046 × 1022
(b) Mass of one proton = 1.67493 × 10–27 kg

Total mass of protons in 34 mg of NH3

= (1.67493 × 10–27 kg) (1.2046 × 1022)

= 2.0176 × 10–5 kg

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Q. 2.3 How many neutrons and protons are there in the following nuclei?

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
Ans.
1. 136C

Atomic mass = 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass) – (Atomic number)

= 13 – 6 = 7

2. NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Atomic mass = 16

Atomic number = 8

Number of protons = 8

Number of neutrons = (Atomic mass) – (Atomic number)

= 16 – 8 = 8

3. NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Atomic mass = 24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass) – (Atomic number)

= 24 – 12 = 12

4.NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) – (Atomic number)

= 56 – 26 = 30

5. NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass) – (Atomic number)

= 88 – 38 = 50

Q. 2.4 Write the complete symbol for the atom with the given atomic number(Z) & Atomic mass(A).

(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9

Ans.

(i) NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

(ii) NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

(iii) NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes) of the yellow light.

Ans. From the expression,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

We get, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes…….. (i)

Where,

ν = frequency of yellow light

c = velocity of light in vacuum = 3 × 108 m/s

λ = wavelength of yellow light = 580 nm = 580 × 10–9 m

Substituting the values in expression (i): NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Thus, frequency of yellow light emitted from the sodium lamp = 5.17 × 1014 s–1

Wave number of yellow light, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.6 Find energy of each of the photons which
(i) correspond to light of frequency 3× 1015 Hz.
(ii) have wavelength of 0.50 Å.

Ans.
(i) Energy (E) of a photon is given by the expression, E = NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where, h = Planck’s constant = 6.626 × 10–34 Js

ν = frequency of light = 3 × 1015 Hz

Substituting the values in the given expression of E:

E = (6.626 × 10–34) (3 × 1015)

E = 1.988 × 10–18 J
(ii)  Energy (E) of a photon having wavelength (λ) is given by the expression,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

h = Planck’s constant = 6.626 × 10–34 Js

c = velocity of light in vacuum = 3 × 108 m/s

Substituting the values in the given expression of E:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.7 Calculate the wavelength,frequency & wave number of a light wave whose period is 2.0×10–10s.
Ans. Frequency (ν) of light   NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Wavelength (λ) of light NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where, c = velocity of light in vacuum = 3×108 m/s

Substituting the value in the given expression of λ:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Wave number NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesof light NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Ans. Energy (E) of a photon = hν

Energy (En) of ‘n’ photons = nhν

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

λ = wavelength of light = 4000 pm = 4000 ×10–12 m

c = velocity of light in vacuum = 3 × 10m/s

h = Planck’s constant = 6.626 × 10–34 Js

Substituting the values in the given expression of n:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 1016.

Q. 2.9 A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate
(i) the energy of the photon (eV),
(ii) the kinetic energy of the emission, and
(iii) the velocity of the photo electron (1 eV= 1.6020 × 10–19 J).
Ans.
(i) Energy (E) of a photon = hν = hc/λ

Where,

h = Planck’s constant = 6.626 × 10–34 Js

c = velocity of light in vacuum = 3 × 10m/s

λ = wavelength of photon = 4 × 10–7 m

Substituting the values in the given expression of E:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, the energy of the photon is 4.97 × 10–19 J.
(ii) The kinetic energy of emission Ek is given by

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= (3.1020 – 2.13) eV = 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.
(iii) The velocity of a photo electron (ν) can be calculated by the expression,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesis the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

v = 5.84 × 105 ms–1

Hence, the velocity of the photoelectron is 5.84 × 105 ms–1.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes AtomQ. 2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.
Ans. Energy of sodium (E) NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 4.947 × 105 J mol–1

= 494.7 × 103 J mol–1 = 494 kJ mol–1

Q. 2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.
Ans. Power of bulb, P = 25 Watt = 25 Js–1

Energy of one photon, E = hν = hc/λ

Substituting the values in the given expression of E:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

E = 34.87 × 10–20 J

Rate of emission of quanta per second

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes) and work function (W0) of the metal.
Ans. Threshold wavelength of radian NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes= 6800 × 10–10 m

Threshold frequency NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesof the metal

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes= 4.41 × 1014 s–1

Thus, the threshold frequency NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesof the metal is 4.41 × 1014 s–1.

Hence, work function (W0) of the metal = hν0

= (6.626 × 10–34 Js) (4.41 × 1014 s–1)

= 2.922 × 10–19 J

Q. 2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with = 4 to an energy level with = 2?
Ans. The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Substituting the values in the given expression of E:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

E = – (4.0875 × 10–19 J)

The negative sign indicates the energy of emission.

Wavelength of light emitted NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Substituting the values in the given expression of λ:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).
Ans. The expression of energy is given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,  

Z = atomic number of the atom

n = principal quantum number

For ionization from n1 = 5 to NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, the energy required for ionization from n = 5 to n = NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesis 8.72 × 10–20 J.

Energy required for n1 = 1 to n = NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

Q. 2.15 What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?
Ans. When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, a total number of (5 4 3 2 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by n(n-1)/2

Given, n = 6

Number of spectral lines =  6(6-1)/2 = 15.

Q. 2.16
(i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Ans.
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

E5 = –8.72 × 10–20 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,

rn = (0.0529 nm) n2

For, n = 5

r5 = (0.0529 nm) (5)2

r5 = 1.3225 nm

Q. 2.17 Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Ans. For the Balmer series, ni = 2. Thus, the expression of wave number NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesis given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Wave number NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesis inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Noteshas to be the smallest.

For NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesto be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3, we get:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.
Ans. Energy (E) of the nth Bohr orbit of an atom is given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

Z = atomic number of the atom

Ground state energy = – 2.18 × 10–11 ergs

= –2.18 × 10–11 × 10–7 J

= – 2.18 × 10–18 J

Energy required to shift the electron from n = 1 to n = 5 is given as: Δ= E5E1

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.19 The electron energy in hydrogen atom is given by En = (–2.18 × 10–18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Ans. Given,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Energy required for ionization from n = 2 is given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

ΔE = 0.545 × 10–18 J  

= 5.45 × 10–19 J

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Here, λ is the longest wavelength causing the transition.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 3647 × 10–10 m  

= 3647 Å

Q. 2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
Ans. According to de Broglie’s equation,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

λ = wavelength of moving particle

= mass of particle

v = velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, the wavelength of the electron moving with a velocity of 2.05 × 107 ms–1 is 3.548 × 10–11 m.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesQ. 2.21 The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
Ans. From De Broglie’s equation,

 λ = h/mv

Given,

Kinetic energy (K.E) of the electron = 3.0 × 10–25 J

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Substituting the value in the expression of λ:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, the wavelength of the electron is 8.9625 × 10–7 m.

Q. 2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?

Na+ , K+ , Mg2+ , Ca2+ , S2–, Ar
Ans. Isoelectronic species have the same number of electrons.

Number of electrons in sodium (Na+) = 11

Number of electrons in (Na+ ) = 10

A positive charge denotes the loss of an electron.

Similarly,

Number of electrons in K+  = 18

Number of electrons in Mg2+  = 10

Number of electrons in Ca2+  = 18

A negative charge denotes the gain of an electron by a species.

Number of electrons in sulphur (S) = 16

∴ Number of electrons in S2- = 18

Number of electrons in argon (Ar) = 18

Hence, the following are isoelectronic species:

1) Na+ and Mg2+  (10 electrons each)

2) K+ , Ca2+ , S2– and Ar (18 electrons each)

Q. 2.23
(i) Write the electronic configurations of the following ions:

(a) H– 
(b) Na+ 
(c) O2–  
(d) F
(ii) What are the atomic numbers of elements whose outermost electrons are represented by:
(a) 3s1  
(b) 2p3 
(c) 3p5?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s1 
(b) [Ne] 3s2 3p3 
(c) [Ar] 4s2 3d1.
Ans.
(i) (a) H ion

The electronic configuration of H atom is 1s1.

A negative charge on the species indicates the gain of an electron by it.

∴ Electronic configuration of H = 1s2

(b) Na+  ion

The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.

A positive charge on the species indicates the loss of an electron by it.

∴ Electronic configuration of Na  = 1s2 2s2 2p6 3s0 or 1s2 2s2 2p6

(c) O2– ion

The electronic configuration of 0 atom is 1s2 2s2 2p4.

A dinegative charge on the species indicates that two electrons are gained by it.

∴ Electronic configuration of O2– ion = 1s2 2s2 p6

(d) F ion

The electronic configuration of F atom is 1s2 2s2 2p5.

A negative charge on the species indicates the gain of an electron by it.

∴ Electron configuration of F ion = 1s2 2s2 2p6

(ii) (a) 3s1

Completing the electron configuration of the element as

1s2 2s2 2p6 3s1.

∴ Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11

(b) 2p3

Completing the electron configuration of the element as

1s2 2s2 2p3.

∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

∴ Atomic number of the element = 7

(c) 3p5

Completing the electron configuration of the element as

1s2 2s2 2p5.

∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

∴ Atomic number of the element = 9

(iii) (a) [He] 2s1

The electronic configuration of the element is [He] 2s1 = 1s2 2s1.

∴ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).

(b) [Ne] 3s2 3p3

The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3.

∴ Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] 3s2 3pis phosphorus (P).

(c) [Ar] 4s2 3d1

The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1.

∴ Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] 4s2 3dis scandium (Sc).


NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
Q. 2.24 What is the lowest value of n that allows g orbitals to exist?
Ans. For g-orbitals, = 4.

As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

∴ For l = 4, minimum value of n = 5


Q. 2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Ans. For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2


Q. 2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce
(i)  the number of protons and
(ii) the electronic configuration of the element.
Ans.
(i) For an atom to be neutral, the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29
(ii) The electronic configuration of the atom is

1s2 2s2 2p6 3s2 3p6 4s2 3d10.

Q. 2.27 Give the number of electrons in the speciesNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, Hand NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
Ans. 

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes:

Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2

∴ Number of electrons in NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes= 2 – 1 = 1

H2:

Number of electrons in H2 = 1 + 1 = 2

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes:

Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16

∴ Number of electrons in NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes= 16 – 1 = 15

Q. 2.28 (i) An atomic orbital has = 3. What are the possible values of l and ml ?

(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible?

1p, 2s, 2p and 3f
Ans.
(i) n = 3 (Given)

For a given value of n, can have values from 0 to (n1).

∴ For n = 3

l = 0, 1, 2

For a given value of l, ml can have (2l  1) values.

For l = 0, m = 0

l = 1, m = – 1, 0, 1

l = 2, m = – 2, – 1, 0, 1, 2

∴ For n = 3

l = 0, 1, 2

m0 = 0

m1 = – 1, 0, 1

m2 = – 2, – 1, 0, 1, 2
(ii) For 3d orbital, l = 2.

For a given value of l, mlcan have (2l  1) values i.e., 5 values.

∴ For = 2

m2 = – 2, – 1, 0, 1, 2
(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3cannot exist.

For p-orbital, l = 1.

For a given value of n, can have values from zero to (n – 1).

∴ For l is equal to 1, the minimum value of n is 2.

Similarly,

For f-orbital, = 4.

For l = 4, the minimum value of is 5.

Hence, 1and 3f do not exist.

Q. 2.29 Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n = 1, = 0,
(b) n = 3; l =1,
(c) = 4; l = 2,
(d) n = 4; l =3.
Ans. (a) n = 1, l = 0 (Given)

The orbital is 1s.

(b) For n = 3 and l = 1

The orbital is 3p.

(c) For n = 4 and l = 2

The orbital is 4d.

(d) For n = 4 and l = 3

The orbital is 4f.

Q. 2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a)   n = 0,  l = 0, ml = 0, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
(b)  n = 1,   l = 0,   ml = 0,
(c)   n = 1,  = 1,  ml = 0, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
(d)  n = 2, l = 1,  ml = 0, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
(e)  = 3,   = 3,   ml = – 3, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
(f)  = 3,  = 1,  ml = 0, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
Ans.
(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible.

For a given value of n, ‘l’ can have values from zero to (n – 1).

For n = 1, l = 0 and not 1.
(d) The given set of quantum numbers is possible.
(e) The given set of quantum numbers is not possible.

For n = 3,

l = 0 to (3 – 1)

l = 0 to 2 i.e., 0, 1, 2
(f) The given set of quantum numbers is possible.

Q. 2.31 How many electrons in an atom may have the following quantum numbers?
(a) n = 4,  NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes
(b) n = 3, l = 0
Ans.
(a) Total number of electrons in an atom for a value of n = 2n2

∴ For n = 4,

Total number of electrons = 2 (4)2 = 32

The given element has a fully filled orbital as

1s2 2s2 2p6 3s2 3p6 4s2 3d10.

Hence, all the electrons are paired.

∴ Number of electrons (having n = 4 and NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes) = 16
(b) = 3, l = 0 indicates that the electrons are present in the 3orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

Q. 2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Ans. Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,  = 1, 2, 3, …

According to de Broglie’s equation:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Substituting the value of ‘mv’ from expression (2) in expression (1):

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Q. 2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He  spectrum?
Ans. For He+ ion, the wave number NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesassociated with the Balmer transition, n = 4 to = 2 is given by:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,  

n1 = 2, n2 = 4

Z = atomic number of helium

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

According to the question, the desired transition for hydrogen will have the same wavelength as that of He+ .

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

By hit and trail method, the equality given by equation (1) is true only when n1 = 1and n2 = 2.

∴ The transition for n2 = 2 to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to = 2 of He+  spectrum.

Q. 2.34 Calculate the energy required for the processNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes The ionization energy for the H atom in the ground state is 2.18 ×10–18 J atom–1
Ans. Energy associated with hydrogen-like species is given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

For the given process,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

An electron is removed from n = 1 to n = ∞.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

∴ The energy required for the process NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Q. 2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.
Ans. 1 m = 100 cm

1 cm = 10–2 m

Length of the scale = 20 cm

= 20 × 10–2 m

Diameter of a carbon atom = 0.15 nm

= 0.15 × 10–9 m

One carbon atom occupies 0.15 × 10–9 m.

∴ Number of carbon atoms that can be placed in a straight line

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.36 (2 × 108) atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Ans. Length of the given arrangement = 2.4 cm

Number of carbon atoms present = 2 × 108

∴ Diameter of carbon atom

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.37 The diameter of zinc atom isNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes.Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Ans.  (a) Radius of zinc atom = diameter/2

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

(b) Length of the arrangement = 1.6 cm

= 1.6 × 10–2 m

Diameter of zinc atom = 2.6 × 10–10 m

∴ Number of zinc atoms present in the arrangement

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.38 A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it.

Ans. Charge on one electron = 1.6022 × 10–19 C

⇒ 1.6022 × 10–19C charge is carried by 1 electron.

∴ Number of electrons carrying a charge of 2.5 × 10–16 C

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it.

Ans:- Charge on the oil drop = 1.282 ×10–18C

Charge on one electron = 1.6022 × 10–19C

∴Number of electrons present on the oil drop

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Ans. A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms. Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of α-particles (positively charged).


Q. 2.41 Symbols NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notescan be written, whereas symbolsNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes are not acceptable. Answer briefly.

Ans. The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence,NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesis acceptable butNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesis not acceptable.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notescan be written but NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notescannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element


Q. 2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans. Let the number of protons in the element be x.

∴ Number of neutrons in the element

= x  + 31.7% of x

= x + 0.317 x

= 1.317 x

According to the question,

Mass number of the element = 81

∴ (Number of protons number of neutrons) = 81

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Hence, the number of protons in the element i.e., x is 35.

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.

∴ The atomic symbol of the element is NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes.


Q. 2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Ans. Let the number of electrons in the ion carrying a negative charge be x.

Then, Number of neutrons present = + 11.1% of x

= x + 0.111 x

= 1.111 x

Number of electrons in the neutral atom = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

∴ Number of protons in the neutral atom = x – 1

Given, Mass number of the ion = 37

∴ (x – 1) + 1.111x = 37

2.111x = 38

x = 18

∴The symbol of the ion is NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Ans. Let the number of electrons present in ion NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

∴ Number of neutrons in it = x + 30.4% of = 1.304 x

Since the ion is tri-positive,

⇒ Number of electrons in neutral atom = x  + 3

∴ Number of protons in neutral atom = x  + 3

Given, Mass number of the ion = 56

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

∴ Number of protons = x + 3 = 23 + 3 = 26

∴ The symbol of the ion NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.45 Arrange the following type of radiations in increasing order of frequency: 

(a) radiation from microwave oven 

(b) amber light from traffic signal 

(c) radiation from FM radio 

(d) cosmic rays from outer space and 

(e) X-rays.

Ans. The increasing order of frequency is as follows:

Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio


Q. 2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.

Ans. Power of laser = Energy with which it emits photons

Power NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

N = number of photons emitted

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy (E):

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 0.3302 × 107 J

= 3.33 × 106 J

Hence, the power of the laser is 3.33 × 106 J.


Q. 2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate

(a) the frequency of emission,
 (b) distance traveled by this radiation in 30 s
 (c) energy of quantum and
 (d) number of quanta present if it produces 2 J of energy.

Ans. Wavelength of radiation emitted = 616 nm

= 616 × 10–9 m (Given)

(a)  Frequency of emission NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression ofNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 4.87 × 108 × 109 × 10–3 s–1

ν = 4.87 × 1014 s–1

Frequency of emission (ν) = 4.87 × 1014 s–1

(b) Velocity of radiation, (c) = 3.0 × 108 ms–1

Distance travelled by this radiation in 30 s

= (3.0 × 108 ms–1) (30 s)

= 9.0 × 109 m

(c) Energy of quantum (E) = hν

(6.626 × 10–34 Js) (4.87 × 1014 s–1)

Energy of quantum (E) = 32.27 × 10–20 J

(d) Energy of one photon (quantum) = 32.27 × 10–20 J

Therefore, 32.27 × 10–20 J of energy is present in 1 quantum.

Number of quanta in 2 J of energy

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 6.19 ×1018

= 6.2 ×1018


Q. 2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

Ans. From the expression of energy of one photon (E),

E = hc / λ

Where,

λ = wavelength of radiation

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of E:

E NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

E = 3.313 × 10–19 J

Energy of one photon = 3.313 × 10–19 J

Number of photons received with 3.15 × 10–18 J energy

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 9.5 ≈ 10


Q. 2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.

Ans. Frequency of radiation (ν),

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

ν = 5.0 × 108 s–1

Energy (E) of source = Nhν

Where,

N = number of photons emitted

h = Planck’s constant

ν = frequency of radiation

Substituting the values in the given expression of (E):

E = (2.5 × 1015) (6.626 × 10–34 Js) (5.0 × 108 s–1)

E = 8.282 × 10–10 J

Hence, the energy of the source (E) is 8.282 × 10–10 J.


Q. 2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Ans. For λ1 = 589 nm

Frequency of transition

Frequency of transition (ν1) = c / λ

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 5.093 × 1014 s–1

Similarly, for λ2 = 589.6 nm

Frequency of transition

Frequency of transition (ν2) = c / λ2 

 NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

=5.088 × 1014 s–1

Energy difference (ΔE) between excited states = E1 – E2

Where,

E2 = energy associated with λ2

E1 = energy associated with λ1

ΔE = hν1 – hν2

= h(ν1ν2)

= (6.626 × 10–34 Js) (5.093 × 1014 – 5.088 × 1014)s–1

= (6.626 × 10–34 J) (5.0 × 10–3 × 1014)

ΔE = 3.31 × 10–22 J.


Q. 2.51 The work function for caesium atom is 1.9 eV. Calculate 

(a) the threshold wavelength and 

(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Ans. It is given that the work function (W0) for caesium atom is 1.9 eV.

(a) From the expression, NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes , we get:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

λ0 = threshold wavelength

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of (λ0):

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes6.53 × 10–7 m

Hence, the threshold wavelengthNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes is 653 nm.

(b) From the expression,NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, we get:

Vo = W0/h

Where, ν0 = threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν0:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

(1 eV = 1.602 × 10–19 J)

ν0 = 4.593 × 1014 s–1

Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014 s–1.

(c) According to the question:

Wavelength used in irradiation (λ) = 500 nm

Kinetic energy = h (ν – ν0)

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 9.3149 × 10–20 J

Kinetic energy of the ejected photoelectron = 9.3149 × 10–20J

Since K.E NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes = 4.52 × 105 ms–1

Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105 ms–1.


Q. 2.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Ans.  
(a) Assuming the threshold wavelength to beNCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes, the kinetic energy of the radiation is given as:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Three different equalities can be formed by the given value as:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Similarly,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Dividing equation (3) by equation (1):

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesThreshold wavelength NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes= 540 nm

 Note: part (b) of the question is not done due to the incorrect values of velocity given in the question.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Ans. From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,

E = W0  K.E

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes⇒ W0 = E – K.E

Energy of incident photon (E)

Where,

c = velocity of radiation

h = Planck’s constant

λ = wavelength of radiation

Substituting the values in the given expression of E:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

E = 4.83 eV

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,

K.E = 0.35 V

K.E= 0.35 eV

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesWork function, W0 = E – K.E

= 4.83 eV – 0.35 eV

= 4.48 eV

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1, calculate the energy with which it is bound to the nucleus.

Ans. Energy of incident photon (E) is given by,

  E = hc / λ

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Energy of the electron ejected (K.E)

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 10.2480 × 10–17 J = 1.025 × 10–16J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

E – K.E = 13.252 × 10–16 J – 1.025 × 10–16 J

= 12.227 × 10–16 J

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as = 3.29 × 1015 (Hz) [1/32 – 1/n2] . Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Ans. Wavelength of transition = 1285 nm

= 1285 × 10–9 m (Given)

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes (Given)

Since  v = c/λ

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

ν = 2.33 × 1014 s–1

Substituting the value of ν in the given expression,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

n = 4.98 ≈ 5

Hence, for the transition to be observed at 1285 nm, n = 5.

The spectrum lies in the infra-red region.


Q. 2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm & ends at 211.6 pm. Name the series to which this transition belongs & the region of the spectrum.

Ans. The radius of the nth orbit of hydrogen-like particles is given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

For radius (r1) = 1.3225 nm

= 1.32225 × 10–9 m

= 1322.25 × 10–12 m

= 1322.25 pm

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Thus, the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series.

Wave number NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesfor the transition is given by,

1.097 × 107 m–1 NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 2.303 × 106 m–1

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesWavelength (λ) associated with the emission transition is given by,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 0.434 ×10–6 m

λ = 434 nm


Q. 2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules & other type of material. If the velocity of the electron in this microscope is 1.6×106 ms–1, calculate de Broglie wavelength associated with this electron.

Ans. From de Broglie’s equation,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 4.55 × 10–10 m = 455 pm

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notesde Broglie’s wavelength associated with the electron is 455 pm.


Q. 2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Ans. From de Broglie’s equation,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

v = velocity of particle (neutron)

h = Planck’s constant

m = mass of particle (neutron)

λ = wavelength

Substituting the values in the expression of velocity (v),

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 4.94 × 102 ms–1

v = 494 ms–1

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesVelocity associated with the neutron = 494 ms–1


Q. 2.59 If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it.

Ans. According to de Broglie’s equation,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

λ = wavelength associated with the electron

h = Planck’s constant

m = mass of electron

v = velocity of electron

Substituting the values in the expression of λ:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

λ = 332 pm

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesWavelength associated with the electron = 332 pm


Q. 2.60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Ans. According to De Broglie’s expression,

λ = h/mv

Substituting the values in the expression,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.61 If the position of the electron is measured within an accuracy of 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.

Ans. From Heisenberg’s uncertainty principle,

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Where,

Δx = uncertainty in position of the electron

Δp = uncertainty in momentum of the electron

Substituting the values in the expression of Δp:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 2.637 × 10–23 Jsm–1

Δp = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s–1)

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev NotesUncertainty in the momentum of the electron = 2.637 × 10–23 kgms–1.

Actual momentum =   NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

= 1.055 × 10–24 kgms–1

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes


Q. 2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

(1). n = 4, l = 2, ml = –2 , ms = –1/2                                         

(2). n = 3, l = 2, ml= 1 , ms = 1/2

(3). n = 4, l = 1, ml = 0 , ms = 1/2                                           

(4). n = 3, l = 2, ml = –2 , ms = –1/2

(5). n = 3, l = 1, ml = –1 , ms= 1/2                                          

(6). n = 4, l = 1, ml = 0 , ms = 1/2

Ans. For = 4 and l = 2, the orbital occupied is 4d.

For = 3 and l = 2, the orbital occupied is 3d.

For = 4 and l = 1, the orbital occupied is 4p.

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.

Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).


Q. 2.63 The bromine atom possesses 35 electrons.It contains 6 electrons in 2p orbital,6 electrons in 3p orbital & 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

Ans. Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases.

Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with ( 35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.


Q. 2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? 

(i) 2s and 3s,
 (ii) 4and 4f, 
 (iii) 3d and 3p

Ans. Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.

(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.

(ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus.

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.


Q. 2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Ans. Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom.The higher the atomic number,the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of ( 14) than aluminium, which has a nuclear charge of ( 13) Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.


Q. 2.66 Indicate the number of unpaired electrons in:

(a) P,                            

(b) Si,                           

(c) Cr,                          

(d) Fe                           

(e) Kr.

Ans. (a) Phosphorus (P):

Atomic number = 15

The electronic configuration of P is:

1s2 2s2 2p6 3s2 3p3

The orbital picture of P can be represented as:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

From the orbital picture, phosphorus has three unpaired electrons.

(b) Silicon (Si):

Atomic number = 14

The electronic configuration of Si is:

1s2 2s2 2p6 3s2 3p2

The orbital picture of Si can be represented as:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

From the orbital picture, silicon has two unpaired electrons.

(c) Chromium (Cr):

Atomic number = 24

The electronic configuration of Cr is:

1s2 2s2 2p6 3s2 3p4s1 3d5

The orbital picture of chromium is:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

From the orbital picture, chromium has six unpaired electrons.

(d) Iron (Fe):

Atomic number = 26

The electronic configuration is:

1s2 2s2 2p6 3s2 3p4s3d6

The orbital picture of chromium is:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

From the orbital picture, iron has four unpaired electrons.

(e) Krypton (Kr):

Atomic number = 36

The electronic configuration is:

1s2 2s2 2p6 3s2 3p4s3d10 4p6

The orbital picture of krypton is:

NCERT Solutions (Part - 1) - Structure of Atom, Class 11, Chemistry | EduRev Notes

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.


Q. 2.67

(a) How many sub-shells are associated with = 4? 

(b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4?

Ans. (a) n = 4 (Given)

For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).

= 0, 1, 2, 3

Thus, four sub-shells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the nth shell = n2

For n = 4

Number of orbitals = 16

If each orbital is taken fully, then it will have 1 electron with ms value of - ½.

∴ Number of electrons with ms value of  (- ½ ) is 16.

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