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All questions of Structure of Atom for Class 11 Exam
Can you explain the answer of this question below:The number of radial nodes for 3p orbital is __________.A:3B:4C:2D:1The answer is d.
|
Krishna Iyer answered |
► Number of radial nodes = n - 1 – 1
► For 3p orbital, n = 3 – 1 – 1 = 1
► Number of radial nodes = 3 – 1 – 1 = 1.
► For 3p orbital, n = 3 – 1 – 1 = 1
► Number of radial nodes = 3 – 1 – 1 = 1.
The nucleus of a tritium atom, 3H, contains- a)three protons
- b)three neutrons
- c)two protons and one neutron
- d)two neutrons and one proton
Correct answer is option 'D'. Can you explain this answer?
The nucleus of a tritium atom, 3H, contains
a)
three protons
b)
three neutrons
c)
two protons and one neutron
d)
two neutrons and one proton
|
Hansa Sharma answered |
Tritium (3H) is a radioactive isotope of hydrogen. The nucleus decays (by emitting an electron and an antineutrino), changing from a triton (one proton and two neutrons) to a 3He nucleus (two protons and one neutron).
Electronic configuration of the element having atomic number 24.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Electronic configuration of the element having atomic number 24.
a)
b)
c)
d)
|
Ananya Sarkar answered |
atomic no 24 is of Cr. Due to half filled orbital stability it doesnot follow Afbau rule so its configuration is
Thomson’s plum pudding model explained:
- a) Existence of electrons
- b)Electrical neutrality of an atom
- c)Existence of atoms
- d)Electrons move in fixed circular orbits
Correct answer is option 'B'. Can you explain this answer?
Thomson’s plum pudding model explained:
a)
Existence of electrons
b)
Electrical neutrality of an atom
c)
Existence of atoms
d)
Electrons move in fixed circular orbits
|
Raghav Bansal answered |
Postulates of Thomson’s atomic model
- An atom consists of a positively charged sphere with electrons filled into it. The negative and positive charge present inside an atom are equal and as a whole, an atom is electrically neutral.
- Thomson’s model of the atom was compared to plum pudding and watermelon. He compared the red edible part of the watermelon to positively charged sphere whereas the seeds of watermelon to negatively charged particles.
One of the orbitals is non-directional- a)s- orbital
- b)f- orbital
- c)d- orbital
- d)p- orbital
Correct answer is option 'A'. Can you explain this answer?
One of the orbitals is non-directional
a)
s- orbital
b)
f- orbital
c)
d- orbital
d)
p- orbital
|
Gaurav Kumar answered |
s orbital is spherically symmetrical hence non directional.
The nucleus of helium contains- a) Four protons
- b) Four neutrons
- c) Four protons and two electrons
- d) Two neutrons and two protons
Correct answer is option 'D'. Can you explain this answer?
The nucleus of helium contains
a)
Four protons
b)
Four neutrons
c)
Four protons and two electrons
d)
Two neutrons and two protons
|
Ayush Joshi answered |
A helium nucleus is made of two protons and two neutrons.
They both have strong interactions.
A helium nucleus has a charge +2.
Energy change associated per mole of atoms with an atomic transition giving rise to radiations of 
What is the value of x?
Correct answer is '4'. Can you explain this answer?
Energy change associated per mole of atoms with an atomic transition giving rise to radiations of What is the value of x?
What is the value of x?
|
Naina Sharma answered |
Energy = n.hv
where,
n = 1 mole or 6.023 × 1023 atoms
h = 6.626×10-34 J.sec
v (frequancy) = 1 Hz
where,
n = 1 mole or 6.023 × 1023 atoms
h = 6.626×10-34 J.sec
v (frequancy) = 1 Hz
⇒ E = 6.023 ×1023 × 6.626×10-34 × 1 = 3.91 × 10-10
so approximately value of x = 4.
so approximately value of x = 4.
Consider the following transitions n = 1 to n = 2 and then, n = 2 to n = 3.
Then- a)
- b)
- c)
- d)
Correct answer is option 'A,C,D'. Can you explain this answer?
Consider the following transitions n = 1 to n = 2 and then, n = 2 to n = 3.
Then
a)
b)
c)
d)
|
Lavanya Menon answered |
Energy values are additives i.e.
Hence, options A, C, D are correct.
Hence, options A, C, D are correct.
Which of the following electronic transitions requires that the greatest quantity of energy be absorbed by a hydrogen atom ?- a)n = 1 to n = 2
- b)n = 2 to n = 4
- c)n = 3 to n = 6
- d)n = 3 to n = ∞
Correct answer is option 'A'. Can you explain this answer?
Which of the following electronic transitions requires that the greatest quantity of energy be absorbed by a hydrogen atom ?
a)
n = 1 to n = 2
b)
n = 2 to n = 4
c)
n = 3 to n = 6
d)
n = 3 to n = ∞
|
|
Naina Sharma answered |
Therefore, electronic transition (a) requires greatest quantity of energy.
Atomic mass of an element is equal to the sum of?- a) Electron and neutron
- b) Electron and proton
- c) Proton and neutron
- d) None of the above
Correct answer is option 'C'. Can you explain this answer?
Atomic mass of an element is equal to the sum of?
a)
Electron and neutron
b)
Electron and proton
c)
Proton and neutron
d)
None of the above
|
Neha Joshi answered |
Atomic Mass is the sum of no. of protons and neutrons.
Direction (Q. Nos. 9-11) This section contains 3 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.Q. Select the correct statement(s).- a)Electromagnetic radiation is a form of energy consisting of oscillating electric and magnetic fields
- b)Visible light is a form of electromagnetic radiation
- c)The electromagnetic spectrum of sunlight received at the Earth's surface differs from that emitted by the Sun
- d)Ultraviolet radiation is absorbed in the stratosphere by ozone and oxygen
Correct answer is option 'A,B,C,D'. Can you explain this answer?
Direction (Q. Nos. 9-11) This section contains 3 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.
Q. Select the correct statement(s).
a)
Electromagnetic radiation is a form of energy consisting of oscillating electric and magnetic fields
b)
Visible light is a form of electromagnetic radiation
c)
The electromagnetic spectrum of sunlight received at the Earth's surface differs from that emitted by the Sun
d)
Ultraviolet radiation is absorbed in the stratosphere by ozone and oxygen
|
|
Geetika Shah answered |
Option A: Oscillations travel through space at a velocity of light, thus electromagnetic radiation is in the form of energy E = hv.
Option B: Radiowaves, microwaves, X-rays, and visible light are part of the Electromagnetic radiation spectrum.
Option C: Some of the radiations are absorbed in the atmosphere, hence the spectrum differs.
Option D: Ozone layer absorbs ultraviolet radiation.
Option B: Radiowaves, microwaves, X-rays, and visible light are part of the Electromagnetic radiation spectrum.
Option C: Some of the radiations are absorbed in the atmosphere, hence the spectrum differs.
Option D: Ozone layer absorbs ultraviolet radiation.
Hence, options A, B, C, and D are correct.
An cation A3+ has 18 electrons. Write the atomic number of A.- a)15
- b)24
- c)12
- d)21
Correct answer is option 'D'. Can you explain this answer?
An cation A3+ has 18 electrons. Write the atomic number of A.
a)
15
b)
24
c)
12
d)
21
|
|
Lavanya Menon answered |
3+ charge means no of electrons will be 3 less than no. of protons. So no. of protons will be 21. So atomic no. is 21.
An element has 18 electrons, and 20 .neutrons. Its charge is —2. What is its mass number?- a)32
- b)38
- c)39
- d) 40
Correct answer is option 'C'. Can you explain this answer?
An element has 18 electrons, and 20 .neutrons. Its charge is —2. What is its mass number?
a)
32
b)
38
c)
39
d)
40
|
|
Rahul Bansal answered |
Atomic mass of element = Number of proton + Number of neutron
Number of proton = Number of electron
It is given that charge on element is -2 it means it is anion.
Atomic mass = 18 + 20 = 38 Answer
Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.Which of the following is correct?- a)1H1 and 2He3 are isotopes
- b)6C14 and 7N14 are isotopes
- c)19K39 and 20Ca40 are isotones
- d)9F19 and 11Na24 are isodiaphers
Correct answer is option 'C'. Can you explain this answer?
Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Which of the following is correct?
a)
1H1 and 2He3 are isotopes
b)
6C14 and 7N14 are isotopes
c)
19K39 and 20Ca40 are isotones
d)
9F19 and 11Na24 are isodiaphers
|
|
Raghav Bansal answered |
Because isotones means the same number of neutron, So, from the question, option c is right, number of neutron in k is,39-19= 20,and,the number of neutron in ca is also 40-20=20 so it is isotones.
where, a and b are constants.
Direction (Q. Nos. 16 - 19) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive)Q. The energy of an electron in the first Bohr orbit of H-atom is -13.6 eV. What is the possible value of quantum number for the excited state to have energy -3.4 eV?
Correct answer is '2'. Can you explain this answer?
Direction (Q. Nos. 16 - 19) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive)
Q. The energy of an electron in the first Bohr orbit of H-atom is -13.6 eV. What is the possible value of quantum number for the excited state to have energy -3.4 eV?
|
Pooja Shah answered |
For Bohr radius;
E = -13.6×Z2/ n2
For H atom, Z=1 and given energy = -3.4
So, we have, -3.4 = -13.6/n2
Or n = 2
E = -13.6×Z2/ n2
For H atom, Z=1 and given energy = -3.4
So, we have, -3.4 = -13.6/n2
Or n = 2
The nature of positive rays depends on?
- a)The nature of discharge tube.
- b)The nature of residual gas.
- c)The nature of electrode.
- d)All of above
Correct answer is option 'B'. Can you explain this answer?
The nature of positive rays depends on?
a)
The nature of discharge tube.
b)
The nature of residual gas.
c)
The nature of electrode.
d)
All of above
|
Om Desai answered |
- The nature of positive rays produced in a vacuum discharge tube depends upon the nature of the gas-filled.
- The positive rays consist of positive ions obtained by removing one or more electrons from gas molecules.
Direction (Q. Nos. 12 and 13) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d)A hypothetical electromagnetic wave is pictured here.
Q. Energy associated with this wave is- a)4.24 x 10-19J
- b)2.12 x 10-19J
- c)1.06 x 10-19J
- d)8.49 x 10-19J
Correct answer is option 'A'. Can you explain this answer?
Direction (Q. Nos. 12 and 13) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d)
A hypothetical electromagnetic wave is pictured here.
Q. Energy associated with this wave is
a)
4.24 x 10-19J
b)
2.12 x 10-19J
c)
1.06 x 10-19J
d)
8.49 x 10-19J
|
|
Hansa Sharma answered |
A to E makes one complete wave.
Values of e/m (charge/mass) in the categories alpha particle (α), electron (e) and protons (p) increase in the order:
- a)α < e < p
- b)p < e < α
- c)e < α < p
- d)α< p < e
Correct answer is option 'D'. Can you explain this answer?
Values of e/m (charge/mass) in the categories alpha particle (α), electron (e) and protons (p) increase in the order:
a)
α < e < p
b)
p < e < α
c)
e < α < p
d)
α< p < e
|
|
Hansa Sharma answered |
E/m values of particle (α), electron (e) and protons (p) increase in the order α< p < e
Which model describes that there is no change in the energy of electrons as long as they keep revolving in the same energy level and atoms remains stable?- a)Rutherford Model
- b)Bohr’s Model
- c)J.J Thomson Model
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Which model describes that there is no change in the energy of electrons as long as they keep revolving in the same energy level and atoms remains stable?
a)
Rutherford Model
b)
Bohr’s Model
c)
J.J Thomson Model
d)
None of the above
|
Suresh Reddy answered |
Bohr Model of atom:
- An atom is made up of three particles: Electrons, neutrons and protons.
- The protons and neutrons are located in a small nucleus at the centre of the atom.
- The electrons revolve rapidly around the nucleus at the centre of the atom.
- There is a limit to the number of electrons that each energy level can hold.
- Each energy level is associated with a fixed amount of energy.
- There is no change in the energy of electrons as long as they keep revolving in the same energy level.
Bohr explained the stability through the concept of revolution of electrons in different energy levels.
The change in the energy of an electron occurs when it jumps from lower to higher energy levels. When it gains energy, it excites from lower to higher and vice versa.
Thus energy is not lost and the atom remains stable.
The change in the energy of an electron occurs when it jumps from lower to higher energy levels. When it gains energy, it excites from lower to higher and vice versa.
Thus energy is not lost and the atom remains stable.
The nature of positive rays depends on?- a) The nature of discharge tube.
- b) The nature of residual gas
- c) All of above.
- d) The nature of electrode
Correct answer is option 'B'. Can you explain this answer?
The nature of positive rays depends on?
a)
The nature of discharge tube.
b)
The nature of residual gas
c)
All of above.
d)
The nature of electrode
|
Abhiram Choudhary answered |
The positive charges in these rays, other than negative cathode rays (which are electrons), depend on the gas that is used because they are cations - atoms with mostly one electron missing and thus one positive charge. So, if you accelerate, argon cations and protons over the same electric potential, the particles in the rays will have the same kinetic energy, but the argon ions will be much slower, as they are much heavier than the protons.
The number of radial nodes for 3p orbital is __________.- a)3
- b)4
- c)2
- d)1
Correct answer is 'D'. Can you explain this answer?
The number of radial nodes for 3p orbital is __________.
a)
3
b)
4
c)
2
d)
1
|
Amrita Kumar answered |
Number of radial nodes = n-1 – 1
For 3p orbital, n = 3 – 1 – 1 = 1
Number of radial nodes = 3 – 1 – 1 = 1.
For 3p orbital, n = 3 – 1 – 1 = 1
Number of radial nodes = 3 – 1 – 1 = 1.
Daltons atomic theory could not explain one of the following- a)Law of multiple Proportions,
- b)Law of conservation of mass
- c)Law of constant composition
- d)Discovery of sub-atomic particles
Correct answer is option 'D'. Can you explain this answer?
Daltons atomic theory could not explain one of the following
a)
Law of multiple Proportions,
b)
Law of conservation of mass
c)
Law of constant composition
d)
Discovery of sub-atomic particles
|
|
Om Desai answered |
Dalton’s atomic theory stated that atoms were indivisible. However, the discovery of subatomic particles (such as protons, electrons, and neutrons) disproved this postulate.
The energy associated with the first orbit in the hydrogen atom is -2.18 x 10−18 J/atom. What is the energy associated with the fifth orbit?
a) -7.72 ×10−20 J/atom
b) -5.72 ×10−20 J/atom
c) -3.72 ×10−20 J/atom
d) -8.72 ×10−20 J/atom
Correct answer is 'D'. Can you explain this answer?
|
|
Geetika Shah answered |
Energy for any n shell = -2.18 x 10−18 )/n'2 (') stands for raise to the power of therefore, E associated with 5 orbit = -2.18 x 10−18/52 = -2.18 × 10-18/25 = -218 × 10-20/25 = -8.72 x 10-20J/atom
Energy of a mole of radio wave photons with a frequency of 909 kHz is- a)6.02 x 10-28 J
- b)3.62 x 10-4 J
- c)1.00 x 10-4 J
- d)6.02 x 10-31 J
Correct answer is option 'B'. Can you explain this answer?
Energy of a mole of radio wave photons with a frequency of 909 kHz is
a)
6.02 x 10-28 J
b)
3.62 x 10-4 J
c)
1.00 x 10-4 J
d)
6.02 x 10-31 J
|
|
Rajeev Nair answered |
E = N0hv
= 6.02 x 1023 x 6.62 x 10-34 Js x 909 x103 s-1
= 3.62 x 10-4 J
= 3.62 x 10-4 J
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms-1 (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)
- a)2.5 nm
- b)0.40 nm
- c) 14.0 nm
- d)32 nm
Correct answer is option 'B'. Can you explain this answer?
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms-1 (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)
a)
2.5 nm
b)
0.40 nm
c)
14.0 nm
d)
32 nm
|
Siddharth Yadav answered |
To calculate the wavelength associated with a proton moving at 1.0 MeV, we can use the de Broglie wavelength equation:
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J*s), and p is the momentum of the proton.
First, let's convert 1.0 MeV to joules:
1 MeV = 1.6 x 10^-13 J
Next, we need to calculate the momentum of the proton. The momentum (p) is given by:
p = sqrt(2 * m * KE)
Where m is the mass of the proton (1.6726219 x 10^-27 kg) and KE is the kinetic energy of the proton (1.0 MeV = 1.6 x 10^-13 J).
Now, let's calculate the momentum:
p = sqrt(2 * (1.6726219 x 10^-27 kg) * (1.6 x 10^-13 J))
p ≈ 1.29 x 10^-19 kg*m/s
Finally, we can calculate the wavelength:
λ = (6.626 x 10^-34 J*s) / (1.29 x 10^-19 kg*m/s)
λ ≈ 5.13 x 10^-16 m
To convert this wavelength to nanometers, we multiply by 10^9:
λ ≈ 5.13 x 10^-16 m * 10^9 nm/m
λ ≈ 5.13 x 10^-7 nm
Therefore, the wavelength associated with a proton moving at 1.0 MeV is approximately 5.13 x 10^-7 nanometers.
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J*s), and p is the momentum of the proton.
First, let's convert 1.0 MeV to joules:
1 MeV = 1.6 x 10^-13 J
Next, we need to calculate the momentum of the proton. The momentum (p) is given by:
p = sqrt(2 * m * KE)
Where m is the mass of the proton (1.6726219 x 10^-27 kg) and KE is the kinetic energy of the proton (1.0 MeV = 1.6 x 10^-13 J).
Now, let's calculate the momentum:
p = sqrt(2 * (1.6726219 x 10^-27 kg) * (1.6 x 10^-13 J))
p ≈ 1.29 x 10^-19 kg*m/s
Finally, we can calculate the wavelength:
λ = (6.626 x 10^-34 J*s) / (1.29 x 10^-19 kg*m/s)
λ ≈ 5.13 x 10^-16 m
To convert this wavelength to nanometers, we multiply by 10^9:
λ ≈ 5.13 x 10^-16 m * 10^9 nm/m
λ ≈ 5.13 x 10^-7 nm
Therefore, the wavelength associated with a proton moving at 1.0 MeV is approximately 5.13 x 10^-7 nanometers.
According to Pauli’s exclusion principle- a)No two electrons in an atom can have the same set of four quantum numbers
- b)Any two electrons in an atom can have the same set of four quantum numbers
- c)All electrons in an atom can have the same spin numbers
- d)Any two electrons in an atom can have the same spin numbers
Correct answer is 'A'. Can you explain this answer?
According to Pauli’s exclusion principle
a)
No two electrons in an atom can have the same set of four quantum numbers
b)
Any two electrons in an atom can have the same set of four quantum numbers
c)
All electrons in an atom can have the same spin numbers
d)
Any two electrons in an atom can have the same spin numbers
|
|
Gaurav Kumar answered |
Paulis exclusion principle stated that no two electron can have same set of all the four quantum nos.
Photoelectric effect established that light- a)behaves like particles
- b)behaves like waves
- c)behaves like rays
- d)behaves like magnetic fields
Correct answer is option 'A'. Can you explain this answer?
Photoelectric effect established that light
a)
behaves like particles
b)
behaves like waves
c)
behaves like rays
d)
behaves like magnetic fields
|
|
Raghav Bansal answered |
Photoelectric effect proves the particle nature of light.
The charge on electron was determined by- a)Crooks
- b)Bohr
- c)Milliken
- d)Schrodinger
Correct answer is option 'C'. Can you explain this answer?
The charge on electron was determined by
a)
Crooks
b)
Bohr
c)
Milliken
d)
Schrodinger
|
|
Naina Bansal answered |
Millikan
The experiment helped earn Millikan a Nobel prize in 1923 but has been a source of some controversy over the years. J. J. Thomson discovered the electron in 1897 when he measured the charge-to-mass ratio for electrons in a beam. But the value of the charge and whether it was fundamental remained open questions.
Find the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit
Correct answer is '3'. Can you explain this answer?
Find the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit
|
|
Neha Sharma answered |
Number of waves = n(n - 1)/2 where n = Principal quantum number or number of orbit number of waves = 3(3 - 1)/2 = 3 * 2/2 = 3
ALTERNATIVE SOLUTIONS :
In general, the number of waves made by a Bohr electron in an orbit is equal to its quantum number.
According to Bohr’s postulate of angular momentum, in the 3rd orbit
Mur = n h/2π
Mur = 3 (h/2π) …..(i) [n = 3]
According to de Broglie relationship
λ = h/mu ….(ii)
Substituting (ii) in (i), we get
(h/λ) r = 3 (h/2π) or 3λ = 2πr
[∵ mu = h/λ]
Thus the circumference of the 3rd orbit is equal to 3 times the wavelength of electron i.e. the electron makes three revolution around the 3rd orbit.
The number of radial nodes for 3p orbital is __________.- a)3
- b)4
- c)2
- d)1
Correct answer is option 'D'. Can you explain this answer?
The number of radial nodes for 3p orbital is __________.
a)
3
b)
4
c)
2
d)
1
|
Swara Saha answered |
Number of radial nodes = n-1 – 1
For 3p orbital, n = 3 – 1 – 1 = 1
Number of radial nodes = 3 – 1 – 1 = 1.
For 3p orbital, n = 3 – 1 – 1 = 1
Number of radial nodes = 3 – 1 – 1 = 1.
As compared to 1s electron of H-atom in ground state, which of the following properties appear(s) in the radial probability density of 2s electron of H-atom in first excited state?- a)Spherical node appear
- b)Electron charge density is highest in the vicinity of the nucleus
- c)Electron density drops to zero after maximum probability is reache
- d)Electron density rises to second highest valu
Correct answer is option 'A,B,C,D'. Can you explain this answer?
As compared to 1s electron of H-atom in ground state, which of the following properties appear(s) in the radial probability density of 2s electron of H-atom in first excited state?
a)
Spherical node appear
b)
Electron charge density is highest in the vicinity of the nucleus
c)
Electron density drops to zero after maximum probability is reache
d)
Electron density rises to second highest valu
|
Jatin Dasgupta answered |
For2s-electron of H-atom (first excited state)
For 1s-electron of H-atom in ground state
At point Ar = a0 r at point Br > a0
(a) True (b) True (c) True (d) True
For 1s-electron of H-atom in ground state
At point Ar = a0 r at point Br > a0
(a) True (b) True (c) True (d) True
Wave number of a spectral line for a given transition is x cm-1 for He+, then its value for Be3+ (isoelectronic of He+)for the same transition is- a)x cm-1
- b)4x cm-1
- c)
- d)2x cm-1
Correct answer is option 'B'. Can you explain this answer?
Wave number of a spectral line for a given transition is x cm-1 for He+, then its value for Be3+ (isoelectronic of He+)for the same transition is
a)
x cm-1
b)
4x cm-1
c)
d)
2x cm-1
|
|
Naina Bansal answered |
The spectral lines of the transitions are directly proportional to the square of atomic number (Z^2).
1/λ = wavenumber = RhZ^2(1/n1^2 −1/n2^2)
For, He^+ , the wave number = x /cm
Then for Be^3+, the wave number will be 4x /cm
Isobars are the atoms with- a)same atomic number but different number of neutrons
- b)same mass number but different atomic number
- c)same atomic number but different mass number
- d)same number of neutrons but different mass number
Correct answer is option 'B'. Can you explain this answer?
Isobars are the atoms with
a)
same atomic number but different number of neutrons
b)
same mass number but different atomic number
c)
same atomic number but different mass number
d)
same number of neutrons but different mass number
|
|
Nandini Patel answered |
Isobars are atoms (nuclides) of different chemical elements that have the same number of nucleons. Correspondingly, isobars differ in atomic number (or number of protons) but have the same mass number.
Which of the following conclusions could not be derived from Rutherford’s α -particle scattering experiment?a) Most of the space in the atom is empty.b) The radius of the atom is about 10^–10 m while that of nucleus is 10^–15 m.c) Electrons move in a circular path of fixed energy called orbits.d) Electrons and the nucleus are held together by electrostatic forces of attraction.Correct answer is 'C'. Can you explain this answer?
|
Preeti Iyer answered |
Conclusions of Rutherford's scattering experiment:
1. Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected.
2. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.
3. A very small fraction of α-particles were deflected by very large angles, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom.
4. From the data, he also calculated that the radius of the nucleus is about 10^5times less than the radius of the atom.
5. Electrons and the nucleus are held together by electrostatic forces of attraction.
In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 ×10−34 Js)- a)1.92 ×10−3m
- b)5.10 ×10−3m
- c)3.84 ×10−3m
- d)1.52 ×10−4m
Correct answer is option 'A'. Can you explain this answer?
In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 ×10−34 Js)
a)
1.92 ×10−3m
b)
5.10 ×10−3m
c)
3.84 ×10−3m
d)
1.52 ×10−4m
|
|
Naina Sharma answered |
Radio frequency region of the electromagnetic spectrum is used for broadcasting. It is- a)Around 106 Hz
- b)Around 1010 Hz
- c)Around 1015 Hz
- d)Around 1013 Hz
Correct answer is 'A'. Can you explain this answer?
Radio frequency region of the electromagnetic spectrum is used for broadcasting. It is
a)
Around 106 Hz
b)
Around 1010 Hz
c)
Around 1015 Hz
d)
Around 1013 Hz
|
|
Hansa Sharma answered |
Radio frequency region is present around 106106 Hz.
Emission spectrum of a material results from the material's (atom or molecules)- a)radiating an inverted spectrum
- b)radiating a continuous spectrum
- c)transition from normal to excited state
- d)transition from excited to normal state
Correct answer is option 'D'. Can you explain this answer?
Emission spectrum of a material results from the material's (atom or molecules)
a)
radiating an inverted spectrum
b)
radiating a continuous spectrum
c)
transition from normal to excited state
d)
transition from excited to normal state
|
|
Pooja Shah answered |
Emission spectrum is recorded when a electron de excites from higher energy level to ground state.
In Lyman series, shortest wavelength of H-atom appears at x m, then longest wavelength in Balmer series of He+ appear at- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In Lyman series, shortest wavelength of H-atom appears at x m, then longest wavelength in Balmer series of He+ appear at
a)
b)
c)
d)
|
Vivek Patel answered |
For the spectral line in H -atom and H -like species (one electron)
For Lyman series
For shortest wavelength (maximum wave number) n2 → ∞
For longest wave length (minimum wave number), n2 = (n1, + 1).
For Balmer series, n1 = 2
For Lyman series
For shortest wavelength (maximum wave number) n2 → ∞
For longest wave length (minimum wave number), n2 = (n1, + 1).
For Balmer series, n1 = 2
Direction (Q. Nos. 12-13) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d)Radius of Bohr’s orbit of H-atom is 52.9 pm. An emission in H-atom starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.Q. Wavelength (in nm) associated with this emission is- a)434.17nm
- b)230.20 nm
- c)144.70nm
- d)289.40 nm
Correct answer is option 'A'. Can you explain this answer?
Direction (Q. Nos. 12-13) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d)
Radius of Bohr’s orbit of H-atom is 52.9 pm. An emission in H-atom starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.
Q.
Wavelength (in nm) associated with this emission is
a)
434.17nm
b)
230.20 nm
c)
144.70nm
d)
289.40 nm
|
|
Satakshi Kumari answered |
Can you explain the answer of this question below:The threshold frequency v0 for a metal is 7.0 x 1014 s-1. Radiation of frequency v = 1.0 x 1015 s-1 hits the metal. Kinetic energy of the emitted electron is
- A:
2.0 x 10-18 J
- B:
1.60 x 10-17 J
- C:
1.60 x 10-19 J
- D:
2.0 x 10-19 J
The answer is d.
The threshold frequency v0 for a metal is 7.0 x 1014 s-1. Radiation of frequency v = 1.0 x 1015 s-1 hits the metal. Kinetic energy of the emitted electron is
2.0 x 10-18 J
1.60 x 10-17 J
1.60 x 10-19 J
2.0 x 10-19 J
|
|
Suresh Reddy answered |
By photoelectric effect
Give the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference between the numbers of total p and total s electrons.
- a)Kr ( atomic no.=36)
- b)Br ( atomic no.=35
- c)Cl ( atomic no.=17)
- d)As ( atomic no.=33)
Correct answer is option 'A'. Can you explain this answer?
Give the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference between the numbers of total p and total s electrons.
a)
Kr ( atomic no.=36)
b)
Br ( atomic no.=35
c)
Cl ( atomic no.=17)
d)
As ( atomic no.=33)
|
Anjana Sharma answered |
It is Krypton .
Electronic configuration: 1s22s22p63s23p64s23d104p6
Number of electrons in s orbital are 8
Number of electrons in p orbital are 18
Number of electrons in d orbitals are 10 (18-8=10)
Electronic configuration: 1s22s22p63s23p64s23d104p6
Number of electrons in s orbital are 8
Number of electrons in p orbital are 18
Number of electrons in d orbitals are 10 (18-8=10)
The electronic configuration 1s22s22p1 belongs to- a)Beryllium
- b)lithium
- c)Boron
- d)carbon
Correct answer is option 'C'. Can you explain this answer?
The electronic configuration 1s22s22p1 belongs to
a)
Beryllium
b)
lithium
c)
Boron
d)
carbon
|
|
Hansa Sharma answered |
It is of boron as it has atomic number 5.
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