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JEE • PHYSICS • ATOMS

Atoms practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 248 questions
1 Medium
1 of 248
Understanding +4 | -1
The de-Broglie wavelength of an electron in the first Bohr orbit is
View answer and explanation
Correct option: D
Equal to the circumference of the first orbit
The correct answer is D (Equal to the circumference of the first orbit) because the de-Broglie wavelength for the first Bohr orbit is equal to the circumference of that orbit.
2 Medium
2 of 248
Remembering +4 | -1
The volume occupied by an atom is greater than the volume of the nucleus by a factor of about
View answer and explanation
Correct option: D
1015{10^{15}}
Option D (1015{10^{15}}) is correct because the ratio of the volume of an atom to the volume of its nucleus is approximately 1015{10^{15}}.
3 Medium
3 of 248
Application +4 | -1
For a hydrogen atom, when electron jump from nn =2=2 to n=1{n}=1, then wave length of radiation emitted found to be λ0\lambda_0. Then, for which transition of electron of a 'He' atom, the wavelength of radiation emitted will be equal to ' λ0\lambda_0 '.
View answer and explanation
Correct option: B
n=4{n}=4 to n=2{n}=2
Option B (n=4{n}=4 to n=2{n}=2) is correct because the wavelength of the emitted radiation matches that of hydrogen when the electron transitions from n=4 to n=2 in He+.
4 Medium
4 of 248
Understanding +4 | -1
Energy of characteristic X-ray is a consequence of
View answer and explanation
Correct option: C
Transition in target atoms
Option C (Transition in target atoms) is correct because characteristic X-rays are produced when an electron transitions between energy levels within the target atoms, releasing energy in the form of X-ray photons.
5 Medium
5 of 248
Application +4 | -1
Taking Rydberg’s constant RH=1.097  ×  107  m{R_H} = 1.097\,\, \times \,\,{10^7}\,\,m, first and second wavelength of Balmer series in hydrogen spectrum is
View answer and explanation
Correct option: D
6552  A0,4863  A06552\,\,{A^0},4863\,\,{A^0}
Option D (6552  A0,4863  A06552\,\,{A^0},4863\,\,{A^0}) is correct because it accurately calculates the first two wavelengths of the Balmer series using the Rydberg formula.
6 Medium
6 of 248
Understanding +4 | -1
When hydrogen atom is in its first excited level, its radius is its ground state radius
View answer and explanation
Correct option: D
Four times
Option D (Four times) is correct because the radius of an orbit is proportional to the square of the principal quantum number.
7 Medium
7 of 248
Analyzing +4 | -1
Which of the following is incorrect, when hydrogen atom is excited from ground state to first excited state?
View answer and explanation
Correct option: D
Its angular momentum decreases by 1.05×10−35Js1.05 \times 10^{-35} \mathrm{Js}
Option D is correct because the angular momentum increases when the hydrogen atom is excited from the ground state to the first excited state, making the statement about decreasing angular momentum incorrect.
8 Medium
8 of 248
Understanding +4 | -1
For which one of the following, Bohr model is not valid ?
View answer and explanation
Correct option: D
Singly ionised neon atom (Ne+)\mathrm{(N}{\mathrm{e}^\mathrm{ + }}\mathrm{)}
Option D (Singly ionised neon atom (Ne+)\mathrm{(N}{\mathrm{e}^\mathrm{ + }}\mathrm{)}) is correct because the Bohr model is only valid for single-electron species, and singly ionized neon has multiple electrons.
9 Medium
9 of 248
Remembering +4 | -1
In Bohr model of the hydrogen atom, the lowest orbit corresponds to
View answer and explanation
Correct option: C
The minimum energy
Option C (The minimum energy) is correct because the lowest orbit (n=1) in the Bohr model corresponds to the minimum energy level, which is -13.6 eV for hydrogen.
10 Medium
10 of 248
Understanding +4 | -1
Assertion (A): Bohr model cannot be extended to two or more electron atoms.

Reason (R): Each electron in the atom interacts not only with the positively charged nucleus but also with all other electrons.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A is correct because the Bohr model's limitations stem from its inability to account for electron-electron interactions in multi-electron atoms.
11 Medium
11 of 248
Application +4 | -1
In Rutherford experiment on α−\alpha -ray scattering the number of particles scattered at 90o{90^\mathrm{o}} be 28 per minute. Then the number of particles scattered per minute by the same foil but are at 60o{60^\mathrm{o}} are.
View answer and explanation
Correct option: B
112
Option B (112) is correct because the number of scattered particles is inversely proportional to sin⁡4(θ/2)\sin^4(\theta/2), and substituting the given angles allows us to find the new number of particles.
12 Medium
12 of 248
Understanding +4 | -1
Consider the following statements concerning electrons :

I. Electrons are universal constituents of mater.

II. J Thomson received the very first Nobel prize in Physics for discovering the electron.

III. The mass of the electron is about 1/2000 of a neutron.

IV. According to Bohr the linear momentum of the electron is quantised in the hydrogen atom. Which of the above statements are not correct?
View answer and explanation
Correct option: D
IV
Option IV is not correct because the angular momentum of the electron is quantized in the hydrogen atom, not the linear momentum.
13 Medium
13 of 248
Analyzing +4 | -1
The highly excited states ( n≫1n \gg 1 ) for hydrogen-like atoms (also called Rydberg states) with nuclear charge ZeZ e are defined by their principal quantum number nn. Which of the following statement(s) is/are false?
View answer and explanation
Correct option: C
The relative change in the energy of two consecutive orbitals varies as 1/n31 / n^3.
Option C is false. The relative change in the energy of two consecutive orbitals varies as 1/n21/n^2, not 1/n31/n^3.
14 Medium
14 of 248
Application +4 | -1
An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength of 45×10−2A.45 \times {10^{ - 2}}\mathrm{A}.

The maximum energy of a photon in the radiation in eV is h=6.62×10−34h = 6.62 \times {10^{ - 34}} Js ,

c=3×108  ms−1)c = 3 \times {10^8}\;\mathrm{m}{\mathrm{s}^{ - 1}})
View answer and explanation
Correct option: A
27500
Option A (27500) is correct because the maximum energy of a photon is inversely proportional to the shortest wavelength, and the calculation yields approximately 27500 eV.
15 Medium
15 of 248
Understanding +4 | -1
An electron with K(eV)K({eV}) collides with hydrogen atom in ground state. The collision is observed to be elastic for
View answer and explanation
Correct option: C
0<K<10.2 eV0 < K < 10.2 \,{eV}
The correct answer is C (0<K<10.2 eV0 < K < 10.2 \,{eV}) because an elastic collision requires that the electron's kinetic energy is not sufficient to excite the hydrogen atom to its first excited state.

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16 Medium
16 of 248
Analyzing +4 | -1
A beam of monochromatic light of wavelength λ\lambda ejects
photoelectrons from a cesium surface (ϕ=1.9 eV).(\phi = 1.9 \text{ eV}). These
photoelectrons are made to collide with hydrogen atoms
in ground state. Find the maximum value of λ\lambda for which

(a) hydrogen atoms may be ionized,
(b) hydrogen atoms
may get excited from the ground state to the first excited
state
View answer and explanation
Correct option: A
(a) 80 nm;80 \mathrm{~nm}; (b) 102 nm;102 \mathrm{~nm};
Option A ((a) 80 nm;80 \mathrm{~nm}; (b) 102 nm;102 \mathrm{~nm};) is correct because it accurately calculates the maximum wavelengths for hydrogen ionization and excitation using the photoelectric effect and energy level transitions.
17 Medium
17 of 248
Remembering +4 | -1
The most penetrating radiation out of the following is
View answer and explanation
Correct option: D
γ\gamma-rays
Option D (γ\gamma-rays) is correct because gamma rays have the highest energy and therefore the greatest penetrating power.
18 Medium
18 of 248
Understanding +4 | -1
Gases begin to conduct electricity at low pressure because
View answer and explanation
Correct option: B
Colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionization of atoms
Option B (Colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionization of atoms) is correct because at low pressure, electrons travel longer distances before colliding, gaining enough energy to ionize atoms.
19 Medium
19 of 248
Application +4 | -1
What is the angular momentum of an electron in Bohr’s hydrogen atom whose energy is – 0.544 eV.
View answer and explanation
Correct option: C
5h2π\frac{{5h}}{{2\pi }}
The correct answer is C (5h2π\frac{{5h}}{{2\pi }}) because the angular momentum of an electron in Bohr's hydrogen atom is given by L=nh2πL = \frac{nh}{2\pi}, where n is the principal quantum number.
20 Medium
20 of 248
Application +4 | -1
For the Bohr’s second orbit of circumference 2πr2\pi r, the de-Broglie wavelength of revolving electron will be
View answer and explanation
Correct option: B
πr\pi r
Option A (πr\pi r) is correct because the de-Broglie wavelength is related to the circumference by 2πr=nλ2 \pi r = n \lambda, and for the second orbit (n=2), λ=πr\lambda = \pi r.
21 Medium
21 of 248
Application +4 | -1
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the ratio of wavelengths is calculated using the Rydberg formula for the Balmer and Lyman series.
22 Medium
22 of 248
Application +4 | -1
If wavelength corresponding to 2nd line of Lyman series is λ\lambda, then the wavelength corresponding to last line of Balmer series will be
View answer and explanation
Correct option: B
32λ9\frac{{32\lambda }}{9}
Option B (32λ9\frac{{32\lambda }}{9}) is correct because it is the calculated wavelength for the last line of the Balmer series based on the given wavelength of the second line of the Lyman series.
23 Medium
23 of 248
Analyzing +4 | -1
An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively. Then
(i) c=1/a\mathrm{c}=1 / \mathrm{a}\quad (ii) a=9/4\mathrm{a}=9 / 4

(iii) b=5/27\mathrm{b}=5 / 27\quad (iv) c=5/27\mathrm{c}=5 / 27

Correct options are -
View answer and explanation
Correct option: A
i, iii, iv
Option A is correct because c = 1/a, b = 5/27, and c = 5/27 are the correct relationships between wavelength, momentum, and energy of the photons emitted.
24 Medium
24 of 248
Understanding +4 | -1
Assertion: The force of repulsion between atomic nucleus and α\alpha-particle varies with distance according to inverse square law.
Reason: Electrons were discovered by J.J. Thomson.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not the correct explanation of the assertion
Option B (If both assertion and reason are true but reason is not the correct explanation of the assertion) is correct because both statements are correct but unrelated.
25 Medium
25 of 248
Remembering +4 | -1
The radius of an electron orbit in a hydrogen atom is of the order of
View answer and explanation
Correct option: C
10−11  m{10^{ - 11}}\,\,m
Option C is correct because the radius of an electron orbit in a hydrogen atom is on the order of 10-11 meters, specifically the Bohr radius.
26 Medium
26 of 248
Understanding +4 | -1
Statement I: Balmer series lies in the visible region of electromagnetic spectrum.
Statement II: 1λ=R(122−1n2)\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), where n=3,4,5,……n=3,4,5, \ldots \ldots.
View answer and explanation
Correct option: A
Statement I is true; Statement II is true; Statement II is the correct explanation for Statement I
Option A (Statement I is true; Statement II is true; Statement II is the correct explanation for Statement I) is correct because the Balmer series, defined by the given formula, falls within the visible region of the electromagnetic spectrum.
27 Medium
27 of 248
Remembering +4 | -1
In the Bohr’s hydrogen atom model, the radius of the stationary orbit is directly proportional to (n=n = principle quantum number)
View answer and explanation
Correct option: D
n2{n^2}
Option D (n2{n^2}) is correct because the radius of the stationary orbit in Bohr's model is directly proportional to the square of the principal quantum number.
28 Medium
28 of 248
Understanding +4 | -1
The wavelength λ\lambda of the Kα{K_\alpha } line of characteristic X-rays spectra varies with atomic number approximately
View answer and explanation
Correct option: C
λ∝1Z2\lambda \propto \frac{1}{{{Z^2}}}
Option C is correct because, according to Moseley's law, the frequency of Kα X-rays is proportional to (Z-1)2, and since wavelength is inversely proportional to frequency, wavelength is inversely proportional to Z2.
29 Medium
29 of 248
Application +4 | -1
Potential energy as a function of rr is given by U=Ar10−Br5,U = \frac{A}{{{r^{10}}}} - \frac{B}{{{r^5}}}, where rr is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be:
View answer and explanation
Correct option: C
(2AB)15{\left( {\frac{{2A}}{B}} \right)^{\frac{1}{5}}}
Option C ((2AB)15{\left( {\frac{{2A}}{B}} \right)^{\frac{1}{5}}}) is correct because at equilibrium, the derivative of the potential energy with respect to r is zero, which leads to this expression for r.
30 Medium
30 of 248
Understanding +4 | -1
Penetrating power of XX-rays does not depend on
View answer and explanation
Correct option: D
Current in the filament
Option D is correct because the penetrating power of X-rays depends on wavelength, energy, and potential difference, but not on the current in the filament.
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Key Takeaways

  • ✓ This set contains 248 single-correct-answer MCQs on ATOMS.
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  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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