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JEE • PHYSICS • ATOMS

Atoms practice · Medium

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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 31-60 of 248 questions
31 Medium
31 of 248
Understanding +4 | -1
If f1,f2{f_1},{f_2} and f3{f_3} are the frequencies of corresponding Kα,Kβ{K_\alpha },{K_\beta } and LαX{L_\alpha }X-rays of an element, then
View answer and explanation
Correct option: C
f2=f1+f3{f_2} = {f_1} + {f_3}
Option C (f2=f1+f3{f_2} = {f_1} + {f_3}) is correct because the energy difference between energy levels corresponds to the frequencies of the emitted X-rays.
32 Medium
32 of 248
Understanding +4 | -1
Solid targets of different elements are bombarded by highly energetic electron beams. The frequency (f)\left( f \right) of the characteristic X-rays emitted from different targets varies with atomic number ZZ as
View answer and explanation
Correct option: B
f∝Z2f \propto {Z^2}
Option B (f∝Z2f \propto {Z^2}) is correct because the frequency of characteristic X-rays is proportional to the square of the atomic number, as described by Moseley's law.
33 Medium
33 of 248
Application +4 | -1
If the wavelength of the first line of the Lyman series for the hydrogen atom is 1210A∘1210\mathop {\text{A}}\limits^ \circ, then what will be the wavelength of the first line of the Balmer series of the hydrogen spectrum?
View answer and explanation
Correct option: A
6434A∘6434\mathop {\text{A}}\limits^ \circ
Option A (6434A∘6434\mathop {\text{A}}\limits^ \circ) is correct because it is calculated using the Rydberg formula for the first line of the Balmer series, given the wavelength of the first Lyman series line.
34 Medium
34 of 248
Application +4 | -1
Which energy state of the triply ionized beryllium has the same electron orbital radius as that of ground state of hydrogen? Given ZZ for Be = 4\mathrm{Be}\,\mathrm{ = }\,\mathrm{4}
View answer and explanation
Correct option: D
n=1n = 1
Option C (n=2n = 2) is correct because the electron orbital radius is proportional to n2/Zn^2/Z, and for the radii to be equal, n2/Zn^2/Z must be the same for both hydrogen and triply ionized beryllium.
35 Medium
35 of 248
Application +4 | -1
The shortest wave length of the spectral line (series limit) of Paschen series is
View answer and explanation
Correct option: A
8204.2  A∘8204.2\,\,\mathop {\text{A}}\limits^ \circ
Option A (8204.2,,A∘8204.2\\,\\,\mathop {\text{A}}\limits^ \circ) is correct because it is calculated using the Rydberg formula for the Paschen series limit (n=3 to n=infinity).
36 Medium
36 of 248
Understanding +4 | -1
Consider the following statements.
A. Atoms of each element emit characteristics spectrum.
B. According to Bohr’s postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit
C. The density of nuclear matter depends on the size of the nucleus
D. A free neutron is stable but a free proton decay is possible
E. Radioactivity is an indication of the instability of nuclei
Choose the correct answer from the options given below.
View answer and explanation
Correct option: B
A, B and E
Option B is correct. Atoms emit characteristic spectra, electrons revolve in stationary orbits, and radioactivity indicates nuclear instability, making A, B, and E correct.
37 Medium
37 of 248
Remembering +4 | -1
The Rydberg constant RR for hydrogen is
View answer and explanation
Correct option: D
R=(14πε0)2.2π2me4ch3R = {\left( {\frac{1}{{4\pi {\varepsilon _0}}}} \right)^2}.\frac{{2{\pi ^2}m{e^4}}}{{c{h^3}}}
Option B (R=(14πε0).2π2me4ch2R = \left( {\frac{1}{{4\pi {\varepsilon _0}}}} \right).\frac{{2{\pi ^2}m{e^4}}}{{c{h^2}}}) is incorrect. The correct formula is R=(14πε0)22π2me4ch3R = {\left( {\frac{1}{{4\pi {\varepsilon _0}}}} \right)^2}\frac{{2{\pi ^2}m{e^4}}}{{c{h^3}}}
38 Medium
38 of 248
Application +4 | -1
If an electron drops from 4th 4^{\text {th }} orbit to 2nd 2^{\text {nd }} orbit in an H{H}-atom, then
View answer and explanation
Correct option: C
It emits a 2.55 eV2.55 \,{eV} photon
The correct answer is that it emits a 2.55 eV photon because the energy difference between the orbits is released as a photon.
39 Medium
39 of 248
Understanding +4 | -1
Statement-I: An electron in hydrogen atom passes fromn =4=4 to n=1n=1 level. The maximum number of photons that can be emitted is 6.

Statement-II: maximum number of photons omitted can only be 4.
View answer and explanation
Correct option: C
Statement-I is true and Statement-II is false
Statement-I is true and Statement-II is false because the maximum number of photons emitted when an electron transitions from n=4 to n=1 is 6, calculated by N = n(n-1)/2.
40 Medium
40 of 248
Understanding +4 | -1
XX-rays are known to be electromagnetic radiations. Therefore the XX-ray photon has
View answer and explanation
Correct option: D
Neither electric charge nor magnetic moment
Option D is correct because X-rays are electromagnetic radiations and thus have neither electric charge nor magnetic moment.
41 Medium
41 of 248
Remembering +4 | -1
Which one of the series of hydrogen spectrum is in the visible region
View answer and explanation
Correct option: B
Balmer series
Option B (Balmer series) is correct because the Balmer series corresponds to transitions to the n=2 energy level, which fall within the visible region of the electromagnetic spectrum.
42 Medium
42 of 248
Application +4 | -1
The energy of an electron in the nthn^{th} orbit is given by En=−13.6/n2eVE_n=-13.6 / n^2 \mathrm{eV}. Calculate the energy required to excite an electron from ground state to the second excited state.
View answer and explanation
Correct option: A
12.09 eV
Option A (12.09 eV) is correct because it is the energy difference between the ground state and the second excited state.
43 Medium
43 of 248
Understanding +4 | -1
How may different wavelengths be observed in the spectrum from a hydrogen sample if atoms are excited to states with quantum number (n)?
View answer and explanation
Correct option: C
n(n−1)2\frac{n(n-1)}{2}
Option C is correct because the number of different wavelengths is given by the formula n(n-1)/2, representing all possible transitions.
44 Medium
44 of 248
Understanding +4 | -1
Which of the following is true?
View answer and explanation
Correct option: C
The Paschan series is a line spectrum in the infrared region
Option C (The Paschan series is a line spectrum in the infrared region) is correct because the Paschen series corresponds to transitions to the n=3 energy level, which emits infrared radiation.
45 Medium
45 of 248
Understanding +4 | -1
X-rays are used in determining the molecular structure of crystalline because its
View answer and explanation
Correct option: C
Its wavelength is comparable to interatomic distance
Option C (Its wavelength is comparable to interatomic distance) is correct because X-rays with wavelengths comparable to atomic spacing are needed for diffraction in crystals.
46 Medium
46 of 248
Application +4 | -1
An α\alpha-particle after passing through a potential difference of VV volts collides with a nucleus. If the atomic number of the nucleus is Z{Z} then the distance of closest approach of α\alpha-particle to the nucleus will be
View answer and explanation
Correct option: B
14.4ZVm14.4 \frac{Z}{V} m
Option B ((cos⁡βcos⁡α)2\left(\frac{\cos \beta}{\cos \alpha}\right)^2) is correct because the ratio of kinetic energies is the square of the ratio of the cosine of the angles, derived from the conservation of momentum in the x-direction.
47 Medium
47 of 248
Understanding +4 | -1
Statement-I: In the experiment of alpha particle scattering, extremely thin gold foils are preferred over other metals.

Statement-II: Gold is a ductile material.
View answer and explanation
Correct option: A
Statement-I and Statement-II are true
Option A is correct because gold is ductile, allowing it to be made into extremely thin foils, which are preferred in alpha particle scattering experiments.
48 Medium
48 of 248
Application +4 | -1
The longest wavelength emitted in hydrogen atom belonging to Balmer series, is
View answer and explanation
Correct option: D
365R\frac{36}{5 R}
Option B (36/5 R) is correct because it is calculated using the Rydberg formula for the Balmer series, where the longest wavelength corresponds to the smallest energy transition (n=3 to n=2).
49 Medium
49 of 248
Understanding +4 | -1
Assertion: When light is passed through a sample of hydrogen atoms in ground state, then wavelengths of absorption lines are same as wavelengths of lines of Lyman series in emission spectrum. and

Reason : In ground state hydrogen atom will absorb only those radiation which will excite to higher energy level.
View answer and explanation
Correct option: A
Assertion is true, Reason is true and Reason is correct explanation for Assertion.
Option A is correct because both the assertion and the reason are true, and the reason correctly explains why absorption lines match Lyman series emission lines.
50 Medium
50 of 248
Understanding +4 | -1
Assertion: The Balmer series lies in the visible region of the electromagnetic spectrum.

Reason : 1λ=R(122−1n2)\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\mathrm{n}^2}\right), where, n=3,4,5,…\mathrm{n}=3,4,5, \ldots
View answer and explanation
Correct option: A
Both Assertion and Reason are true, and Reason is the correct explanation for Assertion
Option A is correct because the Balmer series wavelengths, defined by the given formula, fall within the visible region of the electromagnetic spectrum.
51 Medium
51 of 248
Application +4 | -1
An α\alpha-particle of energy 4 MeV4\,\mathrm{MeV} is scattered by gold foil (Z=79)(Z=79). Calculate the maximum volume in which positive charge of the atom is likely to be concentrated.
View answer and explanation
Correct option: A
7.7×10−40 m7.7 \times 10^{-40} \mathrm{~m}
Option A (7.7×10−40 m7.7 \times 10^{-40} \mathrm{~m}) is correct because it's calculated by finding the distance of closest approach and then using that radius to calculate the volume.
52 Medium
52 of 248
Understanding +4 | -1
Which of the following is true for number of spectral lines in going from Lyman series to PP-fund series
View answer and explanation
Correct option: B
Decreases
Option B (Decreases) is correct because the number of spectral lines decreases as we move from Lyman to P-fund series.
53 Medium
53 of 248
Remembering +4 | -1
In Davisson - Germer experiment maximum intensity is observed at
View answer and explanation
Correct option: B
54∘{54^ \circ } and 50 V
Option B (54∘{54^ \circ } and 50 V) is correct because the maximum intensity is observed at this specific angle and voltage in the Davisson-Germer experiment.
54 Medium
54 of 248
Remembering +4 | -1
Mosley’s law relates the frequencies of line XX-rays with the following characteristics of the target element
View answer and explanation
Correct option: C
Its atomic number
Option C (Its atomic number) is correct because Moseley's law relates the frequency of X-rays to the atomic number of the target element.
55 Medium
55 of 248
Analyzing +4 | -1
Which of the following statements are true regarding Bohr’s model of hydrogen atom

(I) Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus

(II) Radii of allowed orbits of electron are proportional to the principal quantum number

(III) Frequency with which electrons orbit around the nucleus is discrete orbits is inversely proportional to the cube of principal quantum number

(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits

Select correct answer using the codes given below
View answer and explanation
Correct option: A
I and III
Option A (I and III) is correct because orbital speed decreases with increasing orbit radius and frequency is inversely proportional to the cube of the principal quantum number.
56 Medium
56 of 248
Application +4 | -1
X-ray of wavelength λ=2A∘\lambda = 2\mathop A\limits^ \circ is emitted from the metal target. The potential difference applied across the cathode and the metal target is
View answer and explanation
Correct option: C
6200 V
Option C (6200 V) is correct because it is the potential difference required to produce X-rays of the given wavelength, calculated using V=hceλV = \frac{{hc}}{{e\lambda}}.
57 Medium
57 of 248
Application +4 | -1
In an ionisation experiment it is found that a doubly ionised particle enters a magnetic field of 1 T and moves in a circular path of radius 1 m with a speed of 1.6×107  ms−11.6 \times {10^7}\;\mathrm{m}{\mathrm{s}^{ - 1}}. The particle must be
View answer and explanation
Correct option: A
C++{\mathrm{C}^{ + + }}
Option A (C++){\mathrm{C}^{ + + }}) is correct because the mass calculated from the given parameters matches the mass of Carbon-12.
58 Medium
58 of 248
Remembering +4 | -1
Energy of electron in an orbit of HH-atom is
View answer and explanation
Correct option: B
Negative
Option B (Negative) is correct because the energy of an electron in an orbit of a Hydrogen atom is negative, indicating that the electron is bound to the nucleus.
59 Medium
59 of 248
Remembering +4 | -1
In Davisson and Germer experiment intensity of scattered beam of electron was maximum at 50∘50^{\circ} scattering angle and at
View answer and explanation
Correct option: C
54 V54 \mathrm{~V} accelerating potential
Option C ( 54 V54 \mathrm{~V} accelerating potential) is correct because the maximum intensity of the scattered electron beam in the Davisson-Germer experiment occurs at approximately 54V.
60 Medium
60 of 248
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.
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