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JEE • PHYSICS • ATOMS

Atoms practice · Mixed difficulty

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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 61-90 of 257 questions
61 Medium
61 of 257
Application +4 | -1
In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3×10−11 m5.3 \times 10^{-11} \mathrm{~m}. Find the speed of the electron in the ground state. Mass of the electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg} and charge of the electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}.
View answer and explanation
Correct option: D
2.2×106 m/sec2.2 \times 10^6 \mathrm{~m} / \mathrm{sec}
Option D (2.2×106 m/sec2.2 \times 10^6 \mathrm{~m} / \mathrm{sec}) is correct because it is the calculated speed of the electron in the ground state of the hydrogen atom using Coulomb's law and centripetal force.
62 Medium
62 of 257
Understanding +4 | -1
Which of the following statements is/are correct regarding XX-ray tube?
View answer and explanation
Correct option: C
On increasing potential difference between the filament and target, frequency of continuous XX-rays increases
Option C is correct. Increasing the potential difference increases the energy of the electrons hitting the target, thus increasing the frequency of the emitted continuous X-rays.
63 Medium
63 of 257
Understanding +4 | -1
In vacuum an electron of energy 10  keV10\;keV hits tungsten target, then emitted radiation will be
View answer and explanation
Correct option: B
X-rays
Option B (X-rays) is correct because when high-energy electrons strike a metal target in a vacuum, they produce X-rays through bremsstrahlung and characteristic X-ray emission.
64 Medium
64 of 257
Analyzing +4 | -1
Assertion (A): In the Balmer series of hydrogen spectrum as wavelength increases the spectral lines becomes closer.

Reason (R): For higher orbits energy gap between them decreases.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
The correct answer is A because both the assertion and the reason are true, and the reason correctly explains the assertion.
65 Medium
65 of 257
Application +4 | -1
Nth \mathrm{N}^{\text {th }} level of Li2+\mathrm{Li}^{2+} has the same energy as the ground state energy of the hydrogen atom. If rNr_N and r1r_1 is the radii of the Nth \mathrm{N}^{\text {th }} Bohr orbit of Li2+\mathrm{Li}^{2+} and H\mathrm{H} atom respectively, then the ratio rN/r1r_N / r_1 is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the energy of the Nth level of Li2+ equals the ground state energy of hydrogen when n=3, and the ratio r_N/r_1 is then 3.
66 Medium
66 of 257
Understanding +4 | -1
What do you understand by the phrase “ground state atom”?
View answer and explanation
Correct option: A
An unexcited atom
Option A (An unexcited atom) is correct because the ground state of an atom refers to its lowest energy state, where the electrons occupy the lowest possible energy levels.
67 Medium
67 of 257
Analyzing +4 | -1
In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nucleus. The masses and charges of the particles are

particle 1: mass m0{m}_0, charge q0{q}_0

particle 2: mass 2 m02{~m}_0, charge 2q02{q}_0

particle 3: mass 2 m02{~m}_0, charge q0/2{q}_0 / 2

particle 4: mass m0/2{m}_0 / 2, charge 2q02{q}_0

Rank the particles according to the distance of closest approach to the gold nucleus, from smallest to largest.
View answer and explanation
Correct option: A
3, then 1 and 2 tie, then 4
Option A (3, then 1 and 2 tie, then 4) is correct because the distance of closest approach is inversely proportional to the charge-to-mass ratio.
68 Medium
68 of 257
Understanding +4 | -1
Assertion (A): Crystalline solids can cause X-rays to diffract.

Reason (R): Interatomic distance in crystalline solids is of the order of 0.1 nm0.1 \mathrm{~nm}.
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 crystalline solids do diffract X-rays due to the interatomic distances being on the order of X-ray wavelengths, and this distance is the reason for the diffraction.
69 Medium
69 of 257
Understanding +4 | -1
Assertion: For revolving electron, direction of angular momentum and magnetic moment are opposite.
Reason: Charge of electron is negative.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A (If both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because the direction of angular momentum and magnetic moment are opposite for an electron due to its negative charge.
70 Medium
70 of 257
Application +4 | -1
The ratio of the longest to shortest wavelengths in Lyman series of hydrogen spectra is
View answer and explanation
Correct option: D
43\frac{{4}}{3}
Option D (43\frac{{4}}{{3}}) is correct because it's the ratio of the longest wavelength (transition from n=2 to n=1) to the shortest wavelength (transition from infinity to n=1) in the Lyman series.
71 Medium
71 of 257
Understanding +4 | -1
When XX rays pass through a strong uniform magnetic field. Then they
View answer and explanation
Correct option: A
Do not get deflected at all
Option A (Do not get deflected at all) is correct because X-rays are electromagnetic radiation and carry no charge, so they are not affected by magnetic fields.
72 Medium
72 of 257
Understanding +4 | -1
In a Rutherford scattering experiment when a projectile of charge Z1\mathrm{Z}_1 and mass M1\mathrm{M}_1 approaches a target nucleus of charge Z2\mathrm{Z}_2 and mass M2\mathrm{M}_2, the distance of closest approach is r0\mathrm{r}_0. The energy of the projectile is
View answer and explanation
Correct option: A
directly proportional to Z1Z2Z_1 Z_2
Option A is correct because the energy of the projectile is directly proportional to the product of the charges of the projectile and the target nucleus.
73 Medium
73 of 257
Application +4 | -1
Energy levels A,BA, B and CC of a certain atom correspond to increasing values of energy, i.e. EA<EB<ECE_A < E_B < E_C. If λ1,λ2\lambda_1, \lambda_2 and λ3\lambda_3 are the wavelengths of radiations corresponding to transitions CC to BB, BB to AA and CC to AA respectively, then which of the following relations is correct?
View answer and explanation
Correct option: B
λ3−1=λ1−1+λ2−1\lambda_3^{-1}=\lambda_1^{-1}+\lambda_2^{-1}
Option B (λ3−1=λ1−1+λ2−1\lambda_3^{-1}=\lambda_1^{-1}+\lambda_2^{-1}) is correct because it represents the relationship between wavelengths corresponding to energy level transitions.
74 Medium
74 of 257
Understanding +4 | -1
In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is
View answer and explanation
Correct option: C
−1/2\mathrm{ - 1/2}
Option C is correct because the ratio of kinetic energy to potential energy in a Bohr orbit is -1/2.
75 Medium
75 of 257
Understanding +4 | -1
The relation between magnetic moment μ\mu of a revolving electron around the nucleus with principal quantum number nn is
View answer and explanation
Correct option: A
μ∝n\mu \propto n
Option A is correct because the magnetic moment of a revolving electron is directly proportional to the principal quantum number n.
76 Medium
76 of 257
Understanding +4 | -1
Assertion (A): Laser is used to measure distant object as moon.

Reason (R): Lasers are highly coherent source of light.
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 lasers are used for measuring distances due to their high coherence.
77 Easy
77 of 257
Remembering +4 | -1
The value of angular momentum for He+\mathrm{He}^{+} ion in the first Bohr orbit is:
View answer and explanation
Correct option: A
h2π\frac{h}{2 \pi}
Option A (h2π\frac{h}{2 \pi}) is correct because the angular momentum in the first Bohr orbit (n=1) is given by L = nh/2π.
78 Medium
78 of 257
Application +4 | -1
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in
The Balmer series is:
View answer and explanation
Correct option: A
527\frac{5}{{27}}
Option A (527\frac{5}{{27}}) is correct because it is calculated by finding the ratio of the longest wavelengths using the Rydberg formula for the Lyman and Balmer series.
79 Medium
79 of 257
Remembering +4 | -1
The specific charge of an electron is
View answer and explanation
Correct option: C
1.76×1011coulomb/kg1.76 \times {10^{11}}coulomb/kg
Option C (1.76×1011coulomb/kg1.76 \times {10^{11}}coulomb/kg) is correct because it represents the ratio of the electron's charge to its mass, which is the specific charge.
80 Medium
80 of 257
Understanding +4 | -1
In Rutherford scattering experiment, what will be the correct angle for α−\alpha - scattering for an impact parameter b = 0
View answer and explanation
Correct option: D
180∘180^\circ
Option D (180∘180^\circ) is correct because when the impact parameter is zero, the alpha particle approaches the nucleus directly and rebounds back.
81 Medium
81 of 257
Understanding +4 | -1
For which one of the following Bohr model is not valid
View answer and explanation
Correct option: C
Singly ionized neon atom (Ne+)\left({Ne}^{+}\right)
Option C (Singly ionized neon atom (Ne+)\left({Ne}^{+}\right)) is correct because the Bohr model is only valid for single-electron species.
82 Medium
82 of 257
Analyzing +4 | -1
Stopping potentials of 24,100,110,115kV24,100,110,115 \mathrm{kV} are measured for photoelectrons emitted from a certain element when it is irradiated with monochromatic X-rays. The element is used as a target in an X\mathrm{X}-ray tube. The energy of K\mathrm{K} line is -
View answer and explanation
Correct option: B
76KeV76 \mathrm{KeV}
Option B (76KeV76 \mathrm{KeV}) is correct because it represents the difference in binding energies between the K and L shells, which corresponds to the energy of the K line.
83 Medium
83 of 257
Understanding +4 | -1
Consider a photon of continuous X\mathrm{X}-ray and a photon of characteristics X\mathrm{X}-ray of the same wavelength. Which of the following is different for the two photons:
View answer and explanation
Correct option: C
Method of creation
Method of creation is correct because continuous and characteristic X-rays are produced through different mechanisms.
84 Medium
84 of 257
Understanding +4 | -1
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : The beam of electrons show wave nature and exhibit interference and diffraction.
Reason R : Davisson Germer Experimentally verified the wave nature of electrons.
In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: B
Both A and R are correct and R is the correct explanation of A
The correct answer is B (Both A and R are correct and R is the correct explanation of A) because electrons exhibit wave nature, leading to interference and diffraction, and the Davisson-Germer experiment confirmed this wave nature.
85 Medium
85 of 257
Remembering +4 | -1
Calculate the value of 'fine structure constant'.
View answer and explanation
Correct option: A
0.0073
Option A (0.0073) is correct because the fine structure constant is approximately 1/137, which is about 0.0073.
86 Medium
86 of 257
Remembering +4 | -1
Mosley measured the frequency (f)\left( f \right) of the characteristic XX-rays from many metals of different atomic number (Z)\left( \mathrm{Z} \right) and represented his results by a relation known as Mosley’s law. This law is (a, b are constants)
View answer and explanation
Correct option: A
f=a(Z−b)2f = a{\left( {Z - b} \right)^2}
Option A (f=a(Z−b)2f = a{\left( {Z - b} \right)^2}) is correct because Mosley's law relates the frequency of characteristic X-rays to the atomic number of the metal.
87 Medium
87 of 257
Application +4 | -1
Nth \mathrm{N}^{\text {th }} level of Li2+\mathrm{Li}^{2+} has the same energy as the ground state energy of the hydrogen atom. If rNr_N and r1r_1 is the radii of the Nth \mathrm{N}^{\text {th }} Bohr orbit of Li2+\mathrm{Li}^{2+} and H\mathrm{H} atom respectively, then the ratio rN/r1r_N / r_1 is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the energy of the Nth level of Li2+ equals the ground state energy of hydrogen when n=3, and the ratio r_N/r_1 is then 3.
88 Medium
88 of 257
Understanding +4 | -1
STATEMENT-1:If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X\mathrm{X}-rays do not change.

STATEMENT-2:When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X\mathrm{X}-ray energy.
View answer and explanation
Correct option: B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Option B (Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1) is correct because characteristic X-ray wavelengths depend on the target material, not the accelerating potential, and the kinetic energy conversion is a separate fact.
89 Medium
89 of 257
Remembering +4 | -1
The order of the size of nucleus and Bohr radius of an atom respectively are
View answer and explanation
Correct option: A
10−14  m,  10−10  m{10^{ - 14}}\,\,m,\,\,{10^{ - 10}}\,\,m
The correct answer is A (10−14m, 10−10m{10^{ - 14}}\\ m,\,{10^{ - 10}}\\ m) because the approximate size of the nucleus is on the order of 10−1410^{-14} m and the Bohr radius of an atom is on the order of 10−1010^{-10} m.
90 Medium
90 of 257
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.
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