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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 121-150 of 257 questions

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121 Medium
121 of 257
Application +4 | -1
The energy of hydrogen atom in nth orbit is EnE_n, then the energy in nth orbit of single ionised helium atom will be
View answer and explanation
Correct option: A
4En4 E_n
Option A (4En4 E_n) is correct because the energy of an electron in the nth orbit is proportional to Z2Z^2, where Z is the atomic number. For Helium, Z=2, so the energy is 4 times that of Hydrogen.
122 Medium
122 of 257
Remembering +4 | -1
The spectral series of the hydrogen atom that lies in the visible region of the electromagnetic spectrum
View answer and explanation
Correct option: B
Balmer
Option B (Balmer) is correct because the Balmer series corresponds to transitions to the n=2 level, which results in photons in the visible region of the electromagnetic spectrum.
123 Medium
123 of 257
Remembering +4 | -1
The transition of an electron from n2=5,6,……{n_2} = 5,6, \ldots \ldots to n1=4{n_1} = 4 gives rise to
View answer and explanation
Correct option: D
Brackett series
Option D (Brackett series) is correct because the Brackett series corresponds to transitions to the n=4 energy level.
124 Medium
124 of 257
Application +4 | -1
In Bohr model of hydrogen atom, the force
on the electron depends on the principal
quantum number as
View answer and explanation
Correct option: B
F∝1n4F \propto \frac{1}{{{n^4}}}
Option B (F∝1n4F \propto \frac{1}{{{n^4}}}) is correct because the force on the electron in the Bohr model depends on the principal quantum number as F∝1n4F \propto \frac{1}{{{n^4}}}.
125 Medium
125 of 257
Understanding +4 | -1
The spin- orbit interaction has no effect in the
level of the hydrogen atom
View answer and explanation
Correct option: A
s- level
Option A is correct because the spin-orbit interaction depends on the orbital angular momentum, which is zero for s-levels.
126 Medium
126 of 257
Application +4 | -1
The earth revolves round the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation.

(a) Calculate the minimum radius the earth can have for its orbit.

(b) What is the value of the principal quantum number nn for the present radius? Mass of the earth =6.0×1024 kg=6.0 \times 10^{24} \mathrm{~kg}, mass of the sun =2.0×1030 kg=2.0 \times 10^{30} \mathrm{~kg}, earth-sun distance =1.5×1011 m=1.5 \times 10^{11} \mathrm{~m}.
View answer and explanation
Correct option: A
(a) 2.3×10−138 m;2.3 \times 10^{-138} \mathrm{~m}; (b) 2.5×10742.5 \times 10^{74}
Option A ((a) 2.3×10−138 m;2.3 \times 10^{-138} \mathrm{~m}; (b) 2.5×10742.5 \times 10^{74}) is correct because it correctly applies Bohr's quantization rule to the Earth-Sun system and calculates the minimum radius and principal quantum number.
127 Medium
127 of 257
Analyzing +4 | -1
Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?
View answer and explanation
Correct option: A
122 nm122 \mathrm{~nm}
Option A (122 nm122 \mathrm{~nm}) is correct because the Balmer series transitions to n=2, and the subsequent Lyman series transition from n=2 to n=1 corresponds to approximately 122 nm.
128 Medium
128 of 257
Understanding +4 | -1
X-rays are produced in X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X\mathrm{X}-rays has values from
View answer and explanation
Correct option: D
λmin \lambda_{\text {min }} to λmax \lambda_{\text {max }}, where 0<λmin <λmax <∞0<\lambda_{\text {min }}<\lambda_{\text {max }}<\infty
Option D is correct because the continuous X-ray spectrum has a minimum wavelength λmin\lambda_{\text{min}} and a maximum wavelength λmax\lambda_{\text{max}}.
129 Medium
129 of 257
Application +4 | -1
A particle moving with a velocity of 1100th\frac{1}{{100}}\mathrm{th} of that of light will cross a nucleus on about
View answer and explanation
Correct option: D
10−20  s{10^{ - 20}}\,\,\mathrm{s}
Option D (10−20  s{10^{ - 20}}\,\,\mathrm{s}) is correct because the time is calculated by dividing the nuclear distance by the particle's velocity.
130 Medium
130 of 257
Analyzing +4 | -1
Let λα,λβ{\lambda _\alpha },{\lambda _\beta } and λ′α{{\lambda '}_\alpha } denote the wavelengths of the X\mathrm{X}-rays of the Kα,Kβ{K_\alpha },{K_\beta } and Lα{L_\alpha } lines in the characteristic X\mathrm{X}-rays for a metal. Then
View answer and explanation
Correct option: C
1λβ=1λα+1λ′α\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}
Option D (1λα+1λβ=1λα′\frac{1}{{{\lambda _\alpha }}}+\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda ' _\alpha }}}) is incorrect; the correct relation is 1λβ=1λα+1λα′\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{\lambda ' _\alpha }}} based on energy level transitions.
131 Medium
131 of 257
Application +4 | -1
An X-ray tube is operated at 15 kV15 \mathrm{~kV}. Calculate the upper limit of the speed of the electrons striking the target.
View answer and explanation
Correct option: A
7.26×107 m/s7.26 \times 10^7 \mathrm{~m} / \mathrm{s}
Option A (7.26×107 m/s7.26 \times 10^7 \text{{ m/s}}) is correct because it is the maximum speed calculated from the kinetic energy gained by an electron accelerated through the given voltage.
132 Medium
132 of 257
Application +4 | -1
The number of different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to third exited state is
View answer and explanation
Correct option: D
6
Option D (6) is correct because the number of different wavelengths is calculated using the formula n(n-1)/2, where n is the energy level, which in this case is 4.
133 Medium
133 of 257
Application +4 | -1
If the atom 100Fm257_{100}F{m^{257}} follows the Bohr model and radius of 100Fm257_{100}F{m^{257}} is n time the Bohr radius, then find n
View answer and explanation
Correct option: D
14\frac{1}{4}
Option D (14\frac{1}{4}) is correct because the radius of 100Fm257_{100}F{m^{257}} is 14\frac{1}{4} times the Bohr radius, calculated using the Bohr model formula.
134 Medium
134 of 257
Analyzing +4 | -1
In XX-ray tube when the accelerating voltage VV is halved, the difference between the wavelength of Kα{K_\alpha } line and minimum wavelength of continuous XX-ray spectrum
View answer and explanation
Correct option: D
Becomes less than two times
Option D (Becomes less than two times) is correct because when the voltage is halved, the minimum wavelength doubles, but the KαK_\alpha line remains the same, resulting in a difference less than twice the original.
135 Medium
135 of 257
Application +4 | -1
If 10000  V10000\;V is applied across an XX-ray tube, what will be the ratio of de-Broglie wavelength of the incident electrons to the shortest wavelength of XX-ray produced (em\frac{e}{m} for electron is 1.8×1011ckg−11.8 \times {10^{11}}ck{g^{ - 1}})
View answer and explanation
Correct option: B
0.10.1
Option B (0.10.1) is correct because it is the ratio of de-Broglie wavelength to the shortest wavelength of X-ray produced.
136 Medium
136 of 257
Application +4 | -1
Hydrogen atom excites energy level from fundamental state to n =3. Number of spectrum lines according to Bohr is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the number of spectral lines is calculated using the formula N = n(n-1)/2, where n is the energy level.
137 Medium
137 of 257
Analyzing +4 | -1
Considering Bohr's atomic model for hydrogen atom:

(A) the energy of H atom in ground state is same as energy of He+\mathrm{He}^{+} ion in its first excited state.

(B) the energy of H atom in ground state is same as that for Li++\mathrm{Li}^{++} ion in its second excited state.

(C) the energy of H atom in its ground state is same as that of He+\mathrm{He}^{+} ion for its ground state.

(D) the energy of He+\mathrm{He}^{+} ion in its first excited state is same as that for Li++\mathrm{Li}^{++} ion in its ground state.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: B
(A), (B) only
Option B ((A), (B) only) is correct because the energy of H in the ground state is equal to the energy of He+ in its first excited state and Li++ in its second excited state.
138 Medium
138 of 257
Application +4 | -1
The ratio of energies of first two excited states hydrogen atom is
View answer and explanation
Correct option: B
94\frac{9}{4}
Option B (94\frac{9}{4}) is correct because it represents the ratio of the energies of the first two excited states (n=2 and n=3) of the hydrogen atom.
139 Medium
139 of 257
Understanding +4 | -1
In Bohr's model of the hydrogen atom, the pairs of quantities that are quantized among the following is
View answer and explanation
Correct option: C
Energy and angular momentum
Option C is correct because, according to Bohr's model, both energy and angular momentum are quantized, meaning they can only take on discrete values.
140 Medium
140 of 257
Application +4 | -1
Hydrogen atom excites energy level from fundamental state to n =3. Number of spectrum lines according to Bohr is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the number of spectral lines when an electron transitions from n=3 to the ground state is calculated using the formula N = n(n-1)/2.
141 Medium
141 of 257
Application +4 | -1
In the spectrum of single ionised helium, the wavelength of a line observed is almost the same as the first line of Balmer series of hydrogen. It is due to transition of electron.
View answer and explanation
Correct option: A
From n1=6n_1=6 to n2=4n_2=4
Option C (From n1=4n_1=4 to n2=2n_2=2) is incorrect. The correct answer is from n1=6 to n2=4 as the wavelengths are almost the same.
142 Medium
142 of 257
Remembering +4 | -1
The waves emitted when a metal target is bombarded with high energy electrons are
View answer and explanation
Correct option: C
X-rays
Option C (X-rays) is correct because bombarding a metal target with high-energy electrons results in the emission of X-rays.
143 Medium
143 of 257
Remembering +4 | -1
The wavelength of XX-rays is
View answer and explanation
Correct option: B
2  A∘2\;\mathop A\limits^ \circ
Option B (2 Å) is correct because the wavelength of X-rays typically falls within the range of 0.1 Å to 100 Å.
144 Medium
144 of 257
Application +4 | -1
The difference between longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (Atomic number ZZ) equal to Δλ\Delta \lambda. The value of the Rydberg constant for the given atom is:
View answer and explanation
Correct option: C
3151Δλ⋅Z2\frac{31}{5} \frac{1}{\Delta \lambda \cdot Z^2}
Option C (3151Δλ⋅Z2\frac{31}{5} \frac{1}{\Delta \lambda \cdot Z^2}) is correct because it correctly relates the Rydberg constant to the difference in wavelengths of the Balmer and Lyman series.
145 Medium
145 of 257
Application +4 | -1
In XX-ray tube the accelerating potential applied at the anode is VV volt. The minimum wavelength of the emitted x−rayx-r a y will be
View answer and explanation
Correct option: D
hc/eVh c / e V
Option D (hc/eVh c / e V) is correct because it represents the minimum wavelength of emitted X-rays, derived from equating the electron's kinetic energy to the photon's energy.
146 Medium
146 of 257
Analyzing +4 | -1
If the shortest wavelength of Paschen series is given by λ,\lambda, then the longest wavelength of Lyman series will be:
View answer and explanation
Correct option: C
4λ27\frac{4 \lambda}{27}
Option C (4λ27\frac{4 \lambda}{27}) is correct because it correctly relates the shortest wavelength of the Paschen series to the longest wavelength of the Lyman series using the Rydberg formula.
147 Medium
147 of 257
Understanding +4 | -1
Statement-I: The trajectory traced by an incident particle depends on the impact parameter of collisions.

Statement-II: The impact parameter is the perpendicular distance of the initial velocity vector of the incident particle from the centre of the target nucleus.
View answer and explanation
Correct option: A
Statement-I and Statement-II are true
Option A is correct because both statements are true regarding the scattering of alpha particles by a nucleus, where the trajectory depends on the impact parameter.
148 Medium
148 of 257
Application +4 | -1
Maximum wavelength in Balmer series of hydrogen spectrum is (R : Rydberg constant)
View answer and explanation
Correct option: D
36R\frac{{36}}{R}
Option A (4R\frac{4}{R}) is incorrect. The correct answer is 365R\frac{{36}}{{5R}} because it is derived from the Balmer series formula where n1 = 2 and n2 = 3 for the maximum wavelength.
149 Medium
149 of 257
Application +4 | -1
A positive hydrogen like ion having electron at its ground state emits it, if a photon of wavelength 228Ao228 \mathop {\text{A}}\limits^{\text{o}} or less is absorbed by it. Identify the ion.
View answer and explanation
Correct option: B
Z=2Z=2
Option B (Z=2Z=2) is correct because the energy of the emitted photon is related to the atomic number Z of the hydrogen-like ion, and solving for Z gives 2.
150 Medium
150 of 257
Application +4 | -1
Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?
View answer and explanation
Correct option: B
6
The correct answer is 6 because the energy level of single ionized carbon at n=6 has the same energy as the ground state of hydrogen.
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