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All questions of Structure of Atom for JEE Exam
When alpha particles are sent through a thin metal foil, most of them go straight through the foil because : (1984 - 1 Mark)- a)alpha particles are much heavier than electrons
- b)alpha particles are positively charged
- c)most part of the atom is empty space
- d)alpha particle move with high velocity
Correct answer is option 'A,C'. Can you explain this answer?
When alpha particles are sent through a thin metal foil, most of them go straight through the foil because : (1984 - 1 Mark)
a)
alpha particles are much heavier than electrons
b)
alpha particles are positively charged
c)
most part of the atom is empty space
d)
alpha particle move with high velocity
|
Anand Kumar answered |
According to the conclusion of Rutherford alpha-particle scattering experiment.....C is the correct answer .But if it is multiple choice then A is also correct.An alpha particle is just a helium nucleus without any electron -- two protons and two neutrons. It has a much greater mass than betaparticles (electron).
In a hydrogen atom, if energy of an electron in ground state is 13.6. ev, then that in the 2nd excited state is [2002]- a)1.51 eV
- b)3.4 eV
- c)6.04 eV
- d)13.6 eV.
Correct answer is option 'A'. Can you explain this answer?
In a hydrogen atom, if energy of an electron in ground state is 13.6. ev, then that in the 2nd excited state is [2002]
a)
1.51 eV
b)
3.4 eV
c)
6.04 eV
d)
13.6 eV.
|
Raghav Bansal answered |
2nd excited state will be the 3rd energy level.
Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively [2004]- a)16 and 4
- b)12 and 5
- c)12 and 4
- d)16 and 5
Correct answer is option 'B'. Can you explain this answer?
Consider the ground state of Cr atom (X = 24). The number of electrons with the azimuthal quantum numbers, ℓ = 1 and 2 are, respectively [2004]
a)
16 and 4
b)
12 and 5
c)
12 and 4
d)
16 and 5
|
Ananya Das answered |
Electronic configuration of Cr atom (z = 24) = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1
when ℓ = 1, p - subshell,
Numbers of electrons = 12
when ℓ = 2, d - subshell,
Numbers of electrons = 5
when ℓ = 1, p - subshell,
Numbers of electrons = 12
when ℓ = 2, d - subshell,
Numbers of electrons = 5
Bohr model can explain : (1985 - 1 Mark)- a)the spectrum of hydrogen atom only
- b)spectrum of an atom or ion containing one electron only
- c)the spectrum of hydrogen molecule
- d)the solar spectrum
Correct answer is option 'B'. Can you explain this answer?
Bohr model can explain : (1985 - 1 Mark)
a)
the spectrum of hydrogen atom only
b)
spectrum of an atom or ion containing one electron only
c)
the spectrum of hydrogen molecule
d)
the solar spectrum
|
Rohit Jain answered |
Bohr model can explain spectrum of atoms/ions containing one electron only.
The electronic configuration of an element is 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1. This represents its (2000S)- a)excited state
- b)ground state
- c)cationic form
- d)anionic form
Correct answer is option 'B'. Can you explain this answer?
The electronic configuration of an element is 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1. This represents its (2000S)
a)
excited state
b)
ground state
c)
cationic form
d)
anionic form
|
Lavanya Menon answered |
3d 54s1 system is more stable than 3d44s2, hence former is the ground state configuration.
The triad of nuclei that is isotonic is (1988 - 1 Mark)- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The triad of nuclei that is isotonic is (1988 - 1 Mark)
a)
b)
c)
d)
|
|
Parth Patel answered |
Use A----Z IS SANE MEANS DIFFERENCE IS SAME
Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1, accurate upto 0.001% will be [2006](h = 6.63 × 10–34 Js)- a)1.92 × 10–2 m
- b)3.84 × 10–2 m
- c)19.2 × 10–2 m
- d)5.76 × 10–2 m
Correct answer is option 'A'. Can you explain this answer?
Uncertainty in the position of an electron (mass = 9.1 × 10–31 kg) moving with a velocity 300 ms–1, accurate upto 0.001% will be [2006]
(h = 6.63 × 10–34 Js)
a)
1.92 × 10–2 m
b)
3.84 × 10–2 m
c)
19.2 × 10–2 m
d)
5.76 × 10–2 m
|
Ishanya Singh answered |
Mass of electron is 9.1*10^-31 kg=9.1*10^-28gram
Velocity of electron is
300*0.001/100 = 0.003m/s
Uncertainty in position = delta x *m*v= h/4pie
delta x = h/4pie*1/mv
= 6.63*10^-34/4 *3.14*9.1*10^-28*3*10^-2
= 0.0192*10^-4
=1.92*10^-2 m
Ground state electronic configuration of nitrogen atom can be represented by (1999)- a)
- b)
- c)
- d)
Correct answer is option 'A,D'. Can you explain this answer?
Ground state electronic configuration of nitrogen atom can be represented by (1999)
a)
b)
c)
d)
|
|
Rohit Jain answered |
According to Hund's rule pairing of electrons starts only when each of the orbital in a sub shell has one electron each of parallel spin.
∴ (a) and (d) are correct ground state electronic configurations of nitrogen atom in ground state.
The number of d-electrons retained in Fe2+ [2003] (At. no. of Fe = 26) ion is- a)4
- b)5
- c)6
- d)3
Correct answer is option 'C'. Can you explain this answer?
The number of d-electrons retained in Fe2+ [2003] (At. no. of Fe = 26) ion is
a)
4
b)
5
c)
6
d)
3
|
Tanvi Kumar answered |
Explanation:
In order to determine the number of d-electrons retained in the Fe2+ ion, we need to consider the electronic configuration of the neutral Fe atom and the ionization process.
Electronic configuration of neutral Fe atom:
The atomic number of Fe is 26, which means it has 26 electrons. The electronic configuration of Fe is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Ionization of Fe to form Fe2+ ion:
When Fe loses two electrons to form the Fe2+ ion, the electron configuration changes. The electrons are lost from the outermost shell first, which in this case is the 4s orbital.
The electronic configuration of Fe2+ ion is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
Calculating the number of d-electrons retained:
To determine the number of d-electrons retained, we need to count the number of electrons in the 3d orbital after the ionization process.
The Fe2+ ion has a total of 24 electrons (26 - 2). Since the 3d orbital can hold a maximum of 10 electrons, we subtract the remaining electrons in other orbitals from this maximum.
The electron configuration of Fe2+ ion shows that there are 6 electrons in the 3d orbital. Therefore, the number of d-electrons retained in the Fe2+ ion is 6.
Final Answer:
The correct answer is option 'C', which is 6.
In order to determine the number of d-electrons retained in the Fe2+ ion, we need to consider the electronic configuration of the neutral Fe atom and the ionization process.
Electronic configuration of neutral Fe atom:
The atomic number of Fe is 26, which means it has 26 electrons. The electronic configuration of Fe is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
Ionization of Fe to form Fe2+ ion:
When Fe loses two electrons to form the Fe2+ ion, the electron configuration changes. The electrons are lost from the outermost shell first, which in this case is the 4s orbital.
The electronic configuration of Fe2+ ion is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
Calculating the number of d-electrons retained:
To determine the number of d-electrons retained, we need to count the number of electrons in the 3d orbital after the ionization process.
The Fe2+ ion has a total of 24 electrons (26 - 2). Since the 3d orbital can hold a maximum of 10 electrons, we subtract the remaining electrons in other orbitals from this maximum.
The electron configuration of Fe2+ ion shows that there are 6 electrons in the 3d orbital. Therefore, the number of d-electrons retained in the Fe2+ ion is 6.
Final Answer:
The correct answer is option 'C', which is 6.
A 3p orbital has : (1995S)- a)two non spherical nodes
- b)two spherical nodes
- c)one spherical & one non spherical node
- d)one spherical and two non spherical nodes
Correct answer is option 'C'. Can you explain this answer?
A 3p orbital has : (1995S)
a)
two non spherical nodes
b)
two spherical nodes
c)
one spherical & one non spherical node
d)
one spherical and two non spherical nodes
|
Tejas Verma answered |
TIPS/Formulae : Total nodes = n – l
No. of radial nodes = n – l – 1
No. of angular nodes = l
For 3p sub-shell, n = 3, l = 1
∴ No. of radial nodes = n – l – 1 = 3 – 1 – 1 = 1
∴ No. of angular nodes = l = 1
No. of radial nodes = n – l – 1
No. of angular nodes = l
For 3p sub-shell, n = 3, l = 1
∴ No. of radial nodes = n – l – 1 = 3 – 1 – 1 = 1
∴ No. of angular nodes = l = 1
The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [2004]- a)406 nm
- b)192 nm
- c)91 nm
- d)9.1×10–8 nm
Correct answer is option 'C'. Can you explain this answer?
The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m–1) [2004]
a)
406 nm
b)
192 nm
c)
91 nm
d)
9.1×10–8 nm
|
Geetika Shah answered |
TIPS/Formulae :
λ = 91.15 x 10 -9 m ≈ 91nm
Decrease in atomic number is observed during (1998 - 2 Marks)- a)alpha emission
- b)beta emission
- c)positron emission
- d)electron capture.
Correct answer is option 'A,C,D'. Can you explain this answer?
Decrease in atomic number is observed during (1998 - 2 Marks)
a)
alpha emission
b)
beta emission
c)
positron emission
d)
electron capture.
|
|
Gauri Singla answered |
1: Alpha particle is helium nucleus so it's atomic number is 2 since mass number and atomic number is balanced on both sides atomic number of parent nuclei is decreased by 2 2: positron is partcle with positive charge which is equal to magnitude of charge on electron so atomic number of parent nuclei is decreased by 13: electron capture is competitive with positron emission so both process lead to same nuclear transformation
The correct set of quantum numbers for the unpaired electron of chlorine atom is : (1989 - 1 Mark)
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'C'. Can you explain this answer?
The correct set of quantum numbers for the unpaired electron of chlorine atom is : (1989 - 1 Mark)
a)
a
b)
b
c)
c
d)
d
|
|
Sonu Kumar answered |
Last shell is 3p^5
then n( coefficient of P ) = 3 and l=1 for p_orbit al
so correct ans is c
then n( coefficient of P ) = 3 and l=1 for p_orbit al
so correct ans is c
Electromagnetic radiation with maximum wavelength is : (1985 - 1 Mark)- a)ultraviolet
- b)radiowave
- c)X-ray
- d)infrared
Correct answer is option 'B'. Can you explain this answer?
Electromagnetic radiation with maximum wavelength is : (1985 - 1 Mark)
a)
ultraviolet
b)
radiowave
c)
X-ray
d)
infrared
|
Yash Totale answered |
According to order of
wavelength ,radiowave is approximately lie in between 10 ^14 m to 10 cm. they are generated by LC oscillators and used in radio communication.
Which of the following relates to photons both as wave motion and as a stream of particles? (1992 - 1 Mark)- a)Inference
- b)E = mc2
- c)Diffraction
- d)E = hv
Correct answer is option 'D'. Can you explain this answer?
Which of the following relates to photons both as wave motion and as a stream of particles? (1992 - 1 Mark)
a)
Inference
b)
E = mc2
c)
Diffraction
d)
E = hv
|
Sreemoyee Chakraborty answered |
Understanding Photons: Wave-Particle Duality
Photons exhibit both wave-like and particle-like properties, a concept known as wave-particle duality. The correct answer, option 'D' (E = hv), encapsulates this duality effectively.
Wave Motion of Photons
- Photons can behave as waves, demonstrated by phenomena such as diffraction and interference.
- The equation E = hv illustrates that energy (E) of a photon is related to its frequency (v), which is a wave property.
- The wave nature explains how light can spread out, interfere, and exhibit patterns typical of waves.
Particle Nature of Photons
- Photons also behave as discrete packets of energy, or particles.
- The term "quantum" indicates that light exists in individual units or quanta, each with energy given by E = hv.
- This particle aspect is evident in phenomena like the photoelectric effect, where photons knock electrons out of materials.
Conclusion: E = hv as a Bridge
- The equation E = hv serves as a bridge between the wave and particle descriptions of light.
- It indicates that every photon carries a quantifiable amount of energy directly proportional to its frequency.
- This duality is fundamental in modern physics, leading to the development of quantum mechanics.
In summary, option 'D' effectively relates to photons as both waves and particles, making it the correct answer.
Photons exhibit both wave-like and particle-like properties, a concept known as wave-particle duality. The correct answer, option 'D' (E = hv), encapsulates this duality effectively.
Wave Motion of Photons
- Photons can behave as waves, demonstrated by phenomena such as diffraction and interference.
- The equation E = hv illustrates that energy (E) of a photon is related to its frequency (v), which is a wave property.
- The wave nature explains how light can spread out, interfere, and exhibit patterns typical of waves.
Particle Nature of Photons
- Photons also behave as discrete packets of energy, or particles.
- The term "quantum" indicates that light exists in individual units or quanta, each with energy given by E = hv.
- This particle aspect is evident in phenomena like the photoelectric effect, where photons knock electrons out of materials.
Conclusion: E = hv as a Bridge
- The equation E = hv serves as a bridge between the wave and particle descriptions of light.
- It indicates that every photon carries a quantifiable amount of energy directly proportional to its frequency.
- This duality is fundamental in modern physics, leading to the development of quantum mechanics.
In summary, option 'D' effectively relates to photons as both waves and particles, making it the correct answer.
Given that the abundances of isotopes 54Fe, 56Fe and 57 Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is- a)55.85
- b)55.95 (2009S)
- c)55.75
- d)56.05
Correct answer is option 'B'. Can you explain this answer?
Given that the abundances of isotopes 54Fe, 56Fe and 57 Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is
a)
55.85
b)
55.95 (2009S)
c)
55.75
d)
56.05
|
Mahi Choudhary answered |
Given Information:
- Abundances of isotopes:
- 54Fe: 5%
- 56Fe: 90%
- 57Fe: 5%
To find:
The atomic mass of Fe
Solution:
The atomic mass of an element is the weighted average of the masses of its isotopes, taking into account their relative abundances.
Step 1: Find the atomic masses of each isotope
- The atomic mass of 54Fe is 54 amu (atomic mass units).
- The atomic mass of 56Fe is 56 amu.
- The atomic mass of 57Fe is 57 amu.
Step 2: Calculate the weighted average atomic mass
- Multiply the atomic masses of each isotope by their respective abundances:
- (54Fe mass * 0.05) + (56Fe mass * 0.90) + (57Fe mass * 0.05)
- (54 * 0.05) + (56 * 0.90) + (57 * 0.05)
- 2.7 + 50.4 + 2.85
- 55.95 amu
Answer:
Therefore, the atomic mass of Fe is 55.95 amu. Hence, the correct answer is option 'B'.
- Abundances of isotopes:
- 54Fe: 5%
- 56Fe: 90%
- 57Fe: 5%
To find:
The atomic mass of Fe
Solution:
The atomic mass of an element is the weighted average of the masses of its isotopes, taking into account their relative abundances.
Step 1: Find the atomic masses of each isotope
- The atomic mass of 54Fe is 54 amu (atomic mass units).
- The atomic mass of 56Fe is 56 amu.
- The atomic mass of 57Fe is 57 amu.
Step 2: Calculate the weighted average atomic mass
- Multiply the atomic masses of each isotope by their respective abundances:
- (54Fe mass * 0.05) + (56Fe mass * 0.90) + (57Fe mass * 0.05)
- (54 * 0.05) + (56 * 0.90) + (57 * 0.05)
- 2.7 + 50.4 + 2.85
- 55.95 amu
Answer:
Therefore, the atomic mass of Fe is 55.95 amu. Hence, the correct answer is option 'B'.
Many elements have non-integral atomic masses because : (1984 - 1 Mark)- a)they have isotopes
- b)their isotopes have non-integral masses
- c)their isotopes have different masses
- d)the constitutents, neutrons, protons and electrons, combine to give fractional masses
Correct answer is option 'A,C'. Can you explain this answer?
Many elements have non-integral atomic masses because : (1984 - 1 Mark)
a)
they have isotopes
b)
their isotopes have non-integral masses
c)
their isotopes have different masses
d)
the constitutents, neutrons, protons and electrons, combine to give fractional masses
|
|
Anand Kumar answered |
Option B is incorrect because we will see lots of isotopes have integral masses like C,O.
Option D is incorrect because every elements have fundamental particles.
Option A and C are correct.
Reason:- due to isotopes we calculate average mass and that becomes atomic mass of the particular element.
Option D is incorrect because every elements have fundamental particles.
Option A and C are correct.
Reason:- due to isotopes we calculate average mass and that becomes atomic mass of the particular element.
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]Planck’s constant, h = 6.63 × 10–34 Js- a)5 → 2
- b)4 → 1
- c)2 → 5
- d)3 → 2
Correct answer is option 'A'. Can you explain this answer?
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen [2003]
Planck’s constant, h = 6.63 × 10–34 Js
a)
5 → 2
b)
4 → 1
c)
2 → 5
d)
3 → 2
|
Swara Choudhury answered |
The lines falling in the visible region comprise Balmer series. Hence the third line from red would be n1 =2,
n2 = 5 i.e. 5 → 2.
n2 = 5 i.e. 5 → 2.
The number of neutrons in dipositive zinc ion with mass number 70 is (1979)- a)34
- b)36
- c)38
- d)40
Correct answer is option 'D'. Can you explain this answer?
The number of neutrons in dipositive zinc ion with mass number 70 is (1979)
a)
34
b)
36
c)
38
d)
40
|
Mihir Malik answered |
Number of Neutrons in a Dipositive Zinc Ion with Mass Number 70
To determine the number of neutrons in a dipositive zinc ion with mass number 70, we need to understand the composition of the ion and its atomic structure.
1. Understanding the Composition:
- A dipositive ion means that the ion has a charge of +2, indicating the loss of two electrons.
- The element zinc (Zn) has an atomic number of 30, which means it has 30 protons in its nucleus and 30 electrons surrounding the nucleus in its neutral state.
2. Determining the Mass Number:
- The mass number is the sum of the number of protons and neutrons in the nucleus.
- The atomic number of zinc (Zn) is 30, so it has 30 protons.
- To find the number of neutrons, we subtract the atomic number from the mass number.
- In this case, the mass number is given as 70, so the number of neutrons can be calculated as:
Number of neutrons = Mass number - Atomic number
= 70 - 30
= 40
Therefore, the number of neutrons in the dipositive zinc ion with a mass number of 70 is 40. Hence, the correct answer is option 'D' - 40.
To determine the number of neutrons in a dipositive zinc ion with mass number 70, we need to understand the composition of the ion and its atomic structure.
1. Understanding the Composition:
- A dipositive ion means that the ion has a charge of +2, indicating the loss of two electrons.
- The element zinc (Zn) has an atomic number of 30, which means it has 30 protons in its nucleus and 30 electrons surrounding the nucleus in its neutral state.
2. Determining the Mass Number:
- The mass number is the sum of the number of protons and neutrons in the nucleus.
- The atomic number of zinc (Zn) is 30, so it has 30 protons.
- To find the number of neutrons, we subtract the atomic number from the mass number.
- In this case, the mass number is given as 70, so the number of neutrons can be calculated as:
Number of neutrons = Mass number - Atomic number
= 70 - 30
= 40
Therefore, the number of neutrons in the dipositive zinc ion with a mass number of 70 is 40. Hence, the correct answer is option 'D' - 40.
Which of the following does not characterise X-rays? (1992 - 1 Mark)- a)The radiation can ionise gases
- b)It causes ZnS to fluorescence
- c)Deflected by electric and magnetic fields
- d)Have wavelengths shorter than ultraviolet rays
Correct answer is option 'C'. Can you explain this answer?
Which of the following does not characterise X-rays? (1992 - 1 Mark)
a)
The radiation can ionise gases
b)
It causes ZnS to fluorescence
c)
Deflected by electric and magnetic fields
d)
Have wavelengths shorter than ultraviolet rays
|
Sushant Verma answered |
X-rays can ionise gases and cannot get deflected by electric and magnetic fields, wavelength of these rays is 150 to 0.1Å. Thus the wavelength of X-rays is shorter than that of u.v. rays.
Uncertainty in position of a minute particle of mass 25 g in space is 10–5 m. What is the uncertainty in its velocity (in ms–1) ? (h = 6.6 x 10-34 Js) [2002]- a)2.1 x 10–34
- b)0.5 x 10–34
- c)2.1 x 10–28
- d)0.5 x 10–23.
Correct answer is option 'C'. Can you explain this answer?
Uncertainty in position of a minute particle of mass 25 g in space is 10–5 m. What is the uncertainty in its velocity (in ms–1) ? (h = 6.6 x 10-34 Js) [2002]
a)
2.1 x 10–34
b)
0.5 x 10–34
c)
2.1 x 10–28
d)
0.5 x 10–23.
|
Vaibhav Dasgupta answered |
TIPS/Formulae :
= 2.1 x 10-28 ms-1The energy of an electron in the first Bohr orbit of H atom is –13.6 eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are) (1998 - 2 Marks)- a)– 3.4 eV
- b)– 4.2 eV
- c)– 6.8 eV
- d)– 1.5 eV
Correct answer is option 'A,D'. Can you explain this answer?
The energy of an electron in the first Bohr orbit of H atom is –13.6 eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are) (1998 - 2 Marks)
a)
– 3.4 eV
b)
– 4.2 eV
c)
– 6.8 eV
d)
– 1.5 eV
|
Anuj Saini answered |
The energy of an electron on Bohr orbits of hydrogen atoms is given by the expression
Where n takes only integral values. For the first Bohr orbit, n = 1 and it is given that E1 = –13.6 eV
Hence 
of the given values of energy,, only – 3.4 eV and – 1.5 eV can be obtained by substituting n =2 and 3 respectively in the above expression.
of the given values of energy,, only – 3.4 eV and – 1.5 eV can be obtained by substituting n =2 and 3 respectively in the above expression.The ratio of the energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation is : (1986 - 1 Mark)- a)¼
- b)4
- c)½
- d)2
Correct answer is option 'D'. Can you explain this answer?
The ratio of the energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation is : (1986 - 1 Mark)
a)
¼
b)
4
c)
½
d)
2
|
|
Ujjwal Mishra answered |
Actually the energy is inversely proportional to the wavelength so the ratio is
E1/E2=W2/W1
40000/20000
=2
E1/E2=W2/W1
40000/20000
=2
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ( h = 6.6 × 10–34 kg m2s–1, mass of electron, em = 9.1 × 10–31 kg) : [2009]- a)5.10 × 10 –3 m
- b)1.92 × 10 –3 m
- c)3.84 ×10 –3 m
- d)1.52 × 10 –4 m
Correct answer is option 'B'. Can you explain this answer?
In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is ( h = 6.6 × 10–34 kg m2s–1, mass of electron, em = 9.1 × 10–31 kg) : [2009]
a)
5.10 × 10 –3 m
b)
1.92 × 10 –3 m
c)
3.84 ×10 –3 m
d)
1.52 × 10 –4 m
|
|
Juhi Chopra answered |
Given Data:
- Speed of electron (v) = 600 m/s
- Accuracy = 0.005%
- Planck's constant (h) = 6.6 × 10^-34 kg m^2s^-1
- Mass of electron (em) = 9.1 × 10^-31 kg
Finding the Uncertainty in Position:
- The uncertainty principle states that the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) of a particle is greater than or equal to Planck's constant divided by 4π.
- Δx * Δp >= h / 4π
Calculating Uncertainty in Momentum:
- The momentum (p) of an electron is given by mass * velocity.
- Momentum (p) = em * v
- Uncertainty in momentum (Δp) = Accuracy * momentum
- Δp = 0.00005 * em * v
Substitute Values:
- Δp = 0.00005 * 9.1 × 10^-31 kg * 600 m/s
- Δp = 0.0000002725 kg m/s
Calculating Uncertainty in Position:
- Δx * 0.0000002725 >= 6.6 × 10^-34 / (4 * 3.14)
- Δx >= (6.6 × 10^-34) / (4 * 3.14 * 0.0000002725)
- Δx >= 1.922 × 10^-3 m
Therefore, the certainty with which the position of the electron can be located is 1.922 × 10^-3 m, which is option (b).
- Speed of electron (v) = 600 m/s
- Accuracy = 0.005%
- Planck's constant (h) = 6.6 × 10^-34 kg m^2s^-1
- Mass of electron (em) = 9.1 × 10^-31 kg
Finding the Uncertainty in Position:
- The uncertainty principle states that the product of the uncertainty in position (Δx) and the uncertainty in momentum (Δp) of a particle is greater than or equal to Planck's constant divided by 4π.
- Δx * Δp >= h / 4π
Calculating Uncertainty in Momentum:
- The momentum (p) of an electron is given by mass * velocity.
- Momentum (p) = em * v
- Uncertainty in momentum (Δp) = Accuracy * momentum
- Δp = 0.00005 * em * v
Substitute Values:
- Δp = 0.00005 * 9.1 × 10^-31 kg * 600 m/s
- Δp = 0.0000002725 kg m/s
Calculating Uncertainty in Position:
- Δx * 0.0000002725 >= 6.6 × 10^-34 / (4 * 3.14)
- Δx >= (6.6 × 10^-34) / (4 * 3.14 * 0.0000002725)
- Δx >= 1.922 × 10^-3 m
Therefore, the certainty with which the position of the electron can be located is 1.922 × 10^-3 m, which is option (b).
The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1). [2010]- a)594 nm
- b)640 nm
- c)700 nm
- d)494 nm
Correct answer is option 'D'. Can you explain this answer?
The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light capable of breaking a single Cl – Cl bond is (c = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1). [2010]
a)
594 nm
b)
640 nm
c)
700 nm
d)
494 nm
|
|
Anisha Sengupta answered |
Energy required to break one mole of Cl – Cl bonds in Cl2
= 0.4947 × 10–6 m = 494.7 nm
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is [2008]- a)8.51 × 105 J mol–1
- b)6.56 × 105 J mol–1
- c)7.56 × 105 J mol–1
- d)9.84 × 105 J mol–1
Correct answer is option 'D'. Can you explain this answer?
The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol–1. The energy required to excite the electron in the atom from n = 1 to n = 2 is [2008]
a)
8.51 × 105 J mol–1
b)
6.56 × 105 J mol–1
c)
7.56 × 105 J mol–1
d)
9.84 × 105 J mol–1
|
Cstoppers Instructors answered |
(ΔE), The energy required to excite an electron in an atom of hydrogen from n = 1 to n = 2 is ΔE (difference in energy E2 and E1) Values of E2 and E1 are, E1 = – 1.312 × 106 J mol–1
= –3.28 × 105 J mol–1ΔE is given by the relation,
∴ ΔE = E2 – E1 = [–3.28 × 105]– [–1.312 × 106 ] J mol–1
= (–3.28 × 105 + 1.312 × 106) J mol–1
= 9.84 × 105 J mol–1
Thus the correct answer is (d)
If the nitrogen atom has electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s22s22p3, because the electrons would be closer to the nucleus. Yet 1s7 is not observed because it violates.(2002S)- a)Heisenberg uncertainty principle
- b)Hund's rule
- c)Pauli exclusion principle
- d)Bohr postulate of stationary orbits
Correct answer is option 'C'. Can you explain this answer?
If the nitrogen atom has electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s22s22p3, because the electrons would be closer to the nucleus. Yet 1s7 is not observed because it violates.(2002S)
a)
Heisenberg uncertainty principle
b)
Hund's rule
c)
Pauli exclusion principle
d)
Bohr postulate of stationary orbits
|
Nishant Sharma answered |
Correct answer is option C
The outermost electronic configuration of the most electronegative element is (1988 - 1 Mark)- a)ns2 np3
- b)ns2 np4
- c)ns2 np5
- d)ns2 np6
Correct answer is option 'C'. Can you explain this answer?
The outermost electronic configuration of the most electronegative element is (1988 - 1 Mark)
a)
ns2 np3
b)
ns2 np4
c)
ns2 np5
d)
ns2 np6
|
|
Maya Datta answered |
TIPS/Formulae : The element having highest tendency to accept the electron will be most electronegative element.
Configuration ns2, np5 means it requires only one electron to attain nearest noble gas configuration. So, it will be most electronegative element among given choices.
Configuration ns2, np5 means it requires only one electron to attain nearest noble gas configuration. So, it will be most electronegative element among given choices.
Energy of an electron is given by E = – 2.178 × 
Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be : (h = 6.62 × 10 –34 Js and c = 3.0 × 108 ms–1) [JEE M 2013]- a)1.214 × 10–7 m
- b)2.816 × 10.–7 m
- c)6.500 × 10–7 m
- d)8.500 × 10–7 m
Correct answer is option 'A'. Can you explain this answer?
Energy of an electron is given by E = – 2.178 ×
Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be : (h = 6.62 × 10 –34 Js and c = 3.0 × 108 ms–1) [JEE M 2013]
a)
1.214 × 10–7 m
b)
2.816 × 10.–7 m
c)
6.500 × 10–7 m
d)
8.500 × 10–7 m
|
|
Harsh Singhal answered |
E*3/4=hc/l
The principal quantum number of an atom is related to the (1983 - 1 Mark)- a)size of the orbital
- b)spin angular momentum
- c)orbital angular momentum
- d)orientation of the orbital in space
Correct answer is option 'A'. Can you explain this answer?
The principal quantum number of an atom is related to the (1983 - 1 Mark)
a)
size of the orbital
b)
spin angular momentum
c)
orbital angular momentum
d)
orientation of the orbital in space
|
Megha Nair answered |
The principal quantum number (n) is related to the size of the orbital (n = 1, 2, 3.....)
Rutherford’s alpha particle scattering experiment eventually led to the conclusion that : (1986 - 1 Mark)- a)mass and energy are related
- b)electrons occupy space around the nucleus
- c)neutrons are buried deep in the nucleus
- d)th e point of impact with matter can be pr ecisely determined.
Correct answer is option 'B'. Can you explain this answer?
Rutherford’s alpha particle scattering experiment eventually led to the conclusion that : (1986 - 1 Mark)
a)
mass and energy are related
b)
electrons occupy space around the nucleus
c)
neutrons are buried deep in the nucleus
d)
th e point of impact with matter can be pr ecisely determined.
|
Nishtha Kataria answered |
Electrons occupy space around the nucleus
The orbital diagram in which the Aufbau principle is violated is : (1988 - 1 Mark)
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'B'. Can you explain this answer?
The orbital diagram in which the Aufbau principle is violated is : (1988 - 1 Mark)
a)
a
b)
b
c)
c
d)
d
|
Sharmila Chavan answered |
According to Aufbau principle, the orbital of lower energy (2s) should be fully filled before the filling of orbital of higher energy starts.
Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [2004]- a)n = 4, ℓ = 3, m = + 1, s = + ½
- b)n = 4, ℓ = 4, m = – 4, s = – ½
- c)n = 4, ℓ = 3, m = + 4, s = + ½
- d)n = 3, ℓ = 2, m = – 2, s = + ½
Correct answer is option 'A'. Can you explain this answer?
Which of the following sets of quantum numbers is correct for an electron in 4f orbital ? [2004]
a)
n = 4, ℓ = 3, m = + 1, s = + ½
b)
n = 4, ℓ = 4, m = – 4, s = – ½
c)
n = 4, ℓ = 3, m = + 4, s = + ½
d)
n = 3, ℓ = 2, m = – 2, s = + ½
|
Roshni Chavan answered |
The possible quan tum numbers for 4f electron are n = 4, ℓ = 3, m = – 3, –2 –1, 0, 1 , 2 , 3 and s = 
Of various possiblities only option (a) is possible.
Rutherford’s experiment on scattering of a-particles showed for the first time that the atom has (1981 - 1 Mark)- a)electrons
- b)protons
- c)nucleus
- d)neutrons
Correct answer is option 'C'. Can you explain this answer?
Rutherford’s experiment on scattering of a-particles showed for the first time that the atom has (1981 - 1 Mark)
a)
electrons
b)
protons
c)
nucleus
d)
neutrons
|
Nilotpal Goyal answered |
Ernest Rutherford was a New Zealand-born British physicist who is often referred to as the "father of nuclear physics." He is best known for his discovery of the atomic nucleus and for his model of the atom, which proposed that the atom is mostly empty space with a small, dense nucleus at the center.
Rutherford conducted the famous gold foil experiment in 1911, in which he bombarded a thin sheet of gold foil with alpha particles. He expected the alpha particles to pass through the foil with minimal deflection, but to his surprise, some of the particles were deflected at large angles. This led him to conclude that the atom must have a small, positively charged nucleus at its center, surrounded by a cloud of negatively charged electrons.
Rutherford's model of the atom laid the foundation for our understanding of atomic structure and paved the way for further research into nuclear physics. His work also contributed to the development of the first nuclear weapons and nuclear energy.
In addition to his contributions to physics, Rutherford received numerous awards and honors throughout his career, including the Nobel Prize in Chemistry in 1908 for his investigations into the disintegration of the elements and the chemistry of radioactive substances. He was knighted in 1914 and became the first person to be elevated to the peerage as Lord Rutherford of Nelson in 1931.
Rutherford's work and legacy continue to be celebrated in the scientific community, and he is remembered as one of the greatest physicists of the 20th century.
Rutherford conducted the famous gold foil experiment in 1911, in which he bombarded a thin sheet of gold foil with alpha particles. He expected the alpha particles to pass through the foil with minimal deflection, but to his surprise, some of the particles were deflected at large angles. This led him to conclude that the atom must have a small, positively charged nucleus at its center, surrounded by a cloud of negatively charged electrons.
Rutherford's model of the atom laid the foundation for our understanding of atomic structure and paved the way for further research into nuclear physics. His work also contributed to the development of the first nuclear weapons and nuclear energy.
In addition to his contributions to physics, Rutherford received numerous awards and honors throughout his career, including the Nobel Prize in Chemistry in 1908 for his investigations into the disintegration of the elements and the chemistry of radioactive substances. He was knighted in 1914 and became the first person to be elevated to the peerage as Lord Rutherford of Nelson in 1931.
Rutherford's work and legacy continue to be celebrated in the scientific community, and he is remembered as one of the greatest physicists of the 20th century.
The number of nodal planes in a px orbital is (2000S)- a)one
- b)two
- c)three
- d)zero
Correct answer is option 'A'. Can you explain this answer?
The number of nodal planes in a px orbital is (2000S)
a)
one
b)
two
c)
three
d)
zero
|
|
Mohit Mittal answered |
A is correct as the nodal plane is the place where electron finding probability is minimum
The electrons identified by quantum numbers n and l :
(A) n = 4, l = 1
(B) n = 4, l = 0
(C) n = 3, l = 2
(D) n = 3, l = 1
can be placed in order of increasing energy as : [2012]- a)(C) < (D) < (B) < (A)
- b)(D) < (B) < (C) < (A)
- c)(B) < (D) < (A) < (C)
- d)(A) < (C) < (B) < (D)
Correct answer is option 'B'. Can you explain this answer?
The electrons identified by quantum numbers n and l :
(A) n = 4, l = 1
(B) n = 4, l = 0
(C) n = 3, l = 2
(D) n = 3, l = 1
can be placed in order of increasing energy as : [2012]
(A) n = 4, l = 1
(B) n = 4, l = 0
(C) n = 3, l = 2
(D) n = 3, l = 1
can be placed in order of increasing energy as : [2012]
a)
(C) < (D) < (B) < (A)
b)
(D) < (B) < (C) < (A)
c)
(B) < (D) < (A) < (C)
d)
(A) < (C) < (B) < (D)
|
|
Ankurkumar Ankurkumar answered |
The correct ground state electronic configuration of chromium atom is : (1989 - 1 Mark)- a)[Ar] 3d5 4s1
- b)[Ar] 3d4 4s2
- c)[Ar] 3d6 4s0
- d)[Ar] 4d5 4s1
Correct answer is option 'A'. Can you explain this answer?
The correct ground state electronic configuration of chromium atom is : (1989 - 1 Mark)
a)
[Ar] 3d5 4s1
b)
[Ar] 3d4 4s2
c)
[Ar] 3d6 4s0
d)
[Ar] 4d5 4s1
|
Kranti Patil answered |
Option A is correct as in this configuration chromium is in most stable state due to half filled d-orbital and s-orbital...
Which one of the followin g sets of ion s represents the collection of isoelectronic species? [2004](Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)- a)K+, Cl–, Mg2+, Sc3+
- b)Na+, Ca2+, Sc3+, F–
- c)K+, Ca2+, Sc3+, Cl–
- d)Na+, Mg2+, Al3+, Cl–
Correct answer is option 'C'. Can you explain this answer?
Which one of the followin g sets of ion s represents the collection of isoelectronic species? [2004]
(Atomic nos. : F = 9, Cl = 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
a)
K+, Cl–, Mg2+, Sc3+
b)
Na+, Ca2+, Sc3+, F–
c)
K+, Ca2+, Sc3+, Cl–
d)
Na+, Mg2+, Al3+, Cl–
|
|
Arpita Shah answered |
each contains 18 electrons.
The electron s, iden tified by quantum numbers n and l, (i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2, and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as (1999 - 2 Marks)- a)(iv) < (ii) < (iii) < (i)
- b)(ii) < (iv) < (i) < (iii)
- c)(i) < (iii) < (ii) < (iv)
- d)(iii) < (i) < (iv) < (ii)
Correct answer is option 'A'. Can you explain this answer?
The electron s, iden tified by quantum numbers n and l, (i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2, and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as (1999 - 2 Marks)
a)
(iv) < (ii) < (iii) < (i)
b)
(ii) < (iv) < (i) < (iii)
c)
(i) < (iii) < (ii) < (iv)
d)
(iii) < (i) < (iv) < (ii)
|
|
Smruti Sucharita answered |
Here we know that n= principal quantum number..
l= azimuthal quantum number..
so here we can determine the position of electrons..
so 1.4p
2.4s
3.3d
4.3p....
so energy order...
4p>3d>4s>3p
The increasing order (lowest first) for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle (α) is : (1984 - 1 Mark) - a)e, p, n, α
- b)n, p, e, α
- c)n, p, α, e
- d)n, α, p, e
- e), proton (p), neutron (n) and alpha particle (α) is : (1984 - 1 Mark) (a) e, p, n, α (b) n, p, e, α (c) n, p, α, e (d) n, α, p, e
Correct answer is option 'D'. Can you explain this answer?
The increasing order (lowest first) for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle (α) is : (1984 - 1 Mark)
a)
e, p, n, α
b)
n, p, e, α
c)
n, p, α, e
d)
n, α, p, e
e)
, proton (p), neutron (n) and alpha particle (α) is : (1984 - 1 Mark) (a) e, p, n, α (b) n, p, e, α (c) n, p, α, e (d) n, α, p, e
|
|
Mira Yadav answered |
for neutron 
= 0; α-particle = 
= 0.5;proton =
= 1; electron = 
= 1837
= 1; electron = = 1837P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is 4πr2dr. The qualitative sketch of the dependence of P on r is (JEE Adv. 2016)- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is 4πr2dr. The qualitative sketch of the dependence of P on r is (JEE Adv. 2016)
a)
b)
c)
d)
|
|
Rounak Shah answered |
Radial probability function curve for 1s is (D).
Here P is 4πr2R2.
Here P is 4πr2R2.
Rutherford’s scattering experiment is related to the size of the (1983 - 1 Mark)- a)nucleus
- b)atom
- c)electron
- d)neutron
Correct answer is option 'A'. Can you explain this answer?
Rutherford’s scattering experiment is related to the size of the (1983 - 1 Mark)
a)
nucleus
b)
atom
c)
electron
d)
neutron
|
|
Dharam Pal answered |
Because atom have empty space with nucleus to easy find that size of nucleus is small than size of atom
Which one of the following sets of ions represents a collection of isoelectronic species? [2006]- a)N3–, O2–, F–, S2–
- b)Li+, Na+, Mg2+, Ca2+
- c)K+, Cl–, Ca2+, Sc3+
- d)Ba2+, Sr2+, K+, Ca2+
Correct answer is option 'C'. Can you explain this answer?
Which one of the following sets of ions represents a collection of isoelectronic species? [2006]
a)
N3–, O2–, F–, S2–
b)
Li+, Na+, Mg2+, Ca2+
c)
K+, Cl–, Ca2+, Sc3+
d)
Ba2+, Sr2+, K+, Ca2+
|
Raghavendra Ghoshal answered |
(a)
(b) Li+ = 3+1= 4e–, Na+ = 11–1 = 10e–,
Mg++ = 12–2=10e–
Ca++ = 20 – 2 = 18e– (not isoelectronic)
Mg++ = 12–2=10e–
Ca++ = 20 – 2 = 18e– (not isoelectronic)
(c) K+ = 19 – 1= 18e–, Cl– =17 + 1 = 18e–,
Ca++ = 20 – 2 =18e, Sc3+ = 21–3 = 18e–
(isoelectronic)
Ca++ = 20 – 2 =18e, Sc3+ = 21–3 = 18e–
(isoelectronic)
(d) Ba++ 56 – 2 = 54e, Sr++ 38–2 = 36e–
K+= 9–1 = 18e–, Ca++= 20–2 = 18e–
(not isoelectronic)
K+= 9–1 = 18e–, Ca++= 20–2 = 18e–
(not isoelectronic)
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms –1. (Mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10–34 Js)- a)0.40 nm
- b)2.5 nm [2009]
- c)14.0 nm
- d)0.32 nm
Correct answer is option 'A'. Can you explain this answer?
Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms –1. (Mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10–34 Js)
a)
0.40 nm
b)
2.5 nm [2009]
c)
14.0 nm
d)
0.32 nm
|
|
Tanvi Menon answered |
= 3.97 × 10–10 meter = 0.397 nanometer
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