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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 91-120 of 248 questions

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91 Medium
91 of 248
Application +4 | -1
What is the ratio of the circumference of the first Bohr orbit for the electron in the hydrogen atom to the de Broglie wavelength of electrons having the same velocity as the electron in the first Bohr orbit of the hydrogen atom?
View answer and explanation
Correct option: A
1:11: 1
Option A (1: 1) is correct because the circumference of the first Bohr orbit is equal to the de Broglie wavelength of electrons having the same velocity.
92 Medium
92 of 248
Application +4 | -1
The ratio of minimum to maximum wavelength in Balmer series is
View answer and explanation
Correct option: A
0.214583333
Option A (05:09) is correct because it is the ratio of the minimum wavelength (n=3) to the maximum wavelength (n=∞\infty) in the Balmer series.
93 Medium
93 of 248
Understanding +4 | -1
If an electron is moving in the n thn\,th orbit of the hydrogen atom, then its velocity Vn{V_n} for the n thn\,th orbit is given as
View answer and explanation
Correct option: B
vn∝1n{v_n} \propto \frac{1}{n}
Option B (vn∝1n{v_n} \propto \frac{1}{n}) is correct because the velocity of an electron in the nth orbit of a hydrogen atom is inversely proportional to the principal quantum number n.
94 Medium
94 of 248
Application +4 | -1
de Broglie wavelength associated with the orbital electron in nth orbit of hydrogen atom is approximately given by
View answer and explanation
Correct option: C
(3.3n)A∘(3.3 n)\mathop {\text{A}}\limits^ \circ
Option C ((3.3n)A∘(3.3 n)\mathop {\text{A}}\limits^ \circ) is correct because it's the de Broglie wavelength calculated using the circumference of the nth orbit.
95 Medium
95 of 248
Understanding +4 | -1
The absorption transitions between the first and the fourth energy states of hydrogen atom are 3\mathrm{3}. The emission transitions between these states will be
View answer and explanation
Correct option: D
6\mathrm{6}
Option D (6) is correct because the number of emission transitions is calculated using the formula N = n(n-1)/2, where n is the number of energy levels.
96 Medium
96 of 248
Understanding +4 | -1
Assertion: The phenomenon of X-ray production is basically inverse of photoelectric effect.

Reason: X-rays are electromagnetic waves.
View answer and explanation
Correct option: B
If both assertion and reason are true and reason is not correct explanation of assertion.
Option B is correct because both the assertion and reason are true, but the reason doesn't explain the assertion. X-ray production is the inverse of the photoelectric effect, and X-rays are electromagnetic waves.
97 Medium
97 of 248
Remembering +4 | -1
The angular momentum of electron in nth{n^{\mathrm{th}}} orbit is given by
View answer and explanation
Correct option: C
nh2πn\frac{h}{{2\pi }}
Option C (nh2πn\frac{h}{{2\pi }}) is correct because, according to Bohr's quantization condition, the angular momentum of an electron in the nth orbit is an integer multiple of h/2πh/2\pi.
98 Medium
98 of 248
Understanding +4 | -1
Let En=−18ε02n2h2E_n=\frac{-1}{8 \varepsilon_0^2 n^2 h^2} be the energy of the nnth level of H{H}-atom. If all the H{H}-atoms are in the ground state and radiation of frequency (E2−E1)/h\left(E_2-E_1\right) / h falls on it then
View answer and explanation
Correct option: B
Some of atoms will move to the first excited state
Option B is correct because some of the atoms will absorb the radiation and move to the first excited state.
99 Medium
99 of 248
Understanding +4 | -1
XX-rays beam can be deflected by
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because X-rays are electromagnetic radiation and are not deflected by electric or magnetic fields.
100 Medium
100 of 248
Remembering +4 | -1
The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the
View answer and explanation
Correct option: D
Lyman series
Option D is correct. The Lyman series in the hydrogen spectrum lies in the ultraviolet region, corresponding to transitions to the n=1 energy level.
101 Medium
101 of 248
Application +4 | -1
An X-ray machine is operated at 40 kV. The short wavelength limit of continuous X-rays will be (h=6.63×10−34Js,  c  =  3×108  ms−1,  e=1.6×10−19  C)\left( {h = 6.63 \times {{10}^{ - 34}}Js,\;c\; = \;3 \times {{10}^8}\;m{s^{ - 1}},\;e = 1.6 \times {{10}^{ - 19}}\;C} \right)
View answer and explanation
Correct option: A
0.31A∘0.31\mathop A\limits^ \circ
Option A (0.31A∘0.31\mathop A\limits^ \circ) is correct because it is the shortest wavelength calculated using the given voltage and constants.
102 Medium
102 of 248
Application +4 | -1
In which of the following systems will the radius of the first orbit (n=1)\left( {n = 1} \right) be minimum
View answer and explanation
Correct option: D
Doubly ionized lithium
Option D (Doubly ionized lithium) is correct because the radius of the first orbit is inversely proportional to the atomic number Z, and lithium has the highest Z among the options.
103 Medium
103 of 248
Understanding +4 | -1
Number of spectral lines in hydrogen atom is
View answer and explanation
Correct option: D
Infinite
Option D (Infinite) is correct because there are infinitely many possible transitions between energy levels in a hydrogen atom, leading to an infinite number of spectral lines.
104 Medium
104 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.
105 Medium
105 of 248
Application +4 | -1
The angular momentum of electron in nth n^{\text {th }} orbit of H\mathrm{H}-atom is LL. The magnitude of magnetic moment of the electron is (Symbols have their usual meanings)
View answer and explanation
Correct option: D
e2mL\frac{e}{2 m} L
Option D (e2mL\frac{e}{2 m} L) is correct because the magnetic moment is directly proportional to the angular momentum, with the constant of proportionality being e2m\frac{e}{2m}.
106 Medium
106 of 248
Understanding +4 | -1
According to Bohr’s theory the radius of electron in an orbit described by principle quantum number nn and atomic number ZZ is proportional to
View answer and explanation
Correct option: D
n2Z\frac{{{n^2}}}{Z}
Option D is correct because, according to Bohr's theory, the radius of an electron's orbit is proportional to n² divided by Z.
107 Medium
107 of 248
Understanding +4 | -1
Assertion (A): Study of emission spectra of a material can serve as a finger print for identification of the material.

Reason (R): Every element has its characteristic spectrum of emission.
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 assertion and reason are both true, and the reason correctly explains why emission spectra can be used as a fingerprint for material identification.
108 Medium
108 of 248
Understanding +4 | -1
Consider the following two statements A\mathrm{A} and B\mathrm{B} and identify the correct choice in the given answers

A: Line spectra is due to atoms in gaseous state

B: Band spectra is due to molecules
View answer and explanation
Correct option: D
Both A\mathrm{A} and B\mathrm{B} are true
Option D (Both A\mathrm{A} and B\mathrm{B} are true) is correct because line spectra are produced by atoms in the gaseous state, and band spectra are produced by molecules.
109 Medium
109 of 248
Application +4 | -1
In a hypothetical Bohr hydrogen, the mass of the electron is doubled. The energy E0{E_0} and the radius r0{r_0} of the first orbit will be (a0{a_0} is the Bohr radius)
View answer and explanation
Correct option: A
E0=−27.2  eV;  r0=a0/2{E_0} = - 27.2\,\,eV;\,\,{r_0} = {a_0}/2
Option A (E0=−27.2  eV;  r0=a0/2{E_0} = - 27.2\,\,eV;\,\,{r_0} = {a_0}/2) is correct because the energy is directly proportional to the mass and the radius is inversely proportional to the mass of the electron.
110 Medium
110 of 248
Analyzing +4 | -1
Assertion (A): A laser beam of 0.2 W0.2 \mathrm{~W} can drill holes through a metal sheet whereas a 1000 W1000 \mathrm{~W} torch light cannot.

Reason (R): The frequency of laser light in much higher than that of torch light.
View answer and explanation
Correct option: C
(A) is true but (R) is false
Option C ((A) is true but (R) is false) is correct because a laser can drill holes due to its high intensity and focused beam, not its frequency.
111 Medium
111 of 248
Understanding +4 | -1
If an electron oscillates at a frequency of 1  GHz1\;GHz it gives
View answer and explanation
Correct option: D
None of these
Option D is correct because an electron oscillating at 1 GHz corresponds to radio waves, not X-rays, microwaves, or infrared rays, and the question is about radiation due to oscillation, not atomic transitions.
112 Medium
112 of 248
Understanding +4 | -1
Minimum wavelength of XX-ray produced depends on accelerating potential VV between cathode and anode as
View answer and explanation
Correct option: D
λmin⁡∝1V\lambda_{\min } \propto \frac{1}{V}
Option D (λmin⁡∝1V\lambda_{\min } \propto \frac{1}{V}) is correct because the minimum wavelength of X-rays produced is inversely proportional to the accelerating potential V.
113 Medium
113 of 248
Understanding +4 | -1
The angular momentum of an electron in a hydrogen atom is proportional to
View answer and explanation
Correct option: C
r\sqrt{r}
Option C is correct because the angular momentum of an electron in a hydrogen atom is quantized and proportional to the square root of the radius.
114 Medium
114 of 248
Application +4 | -1
What is the wavelength of the most energetic photon emitted in the Balmer series of the hydrogen atom?
View answer and explanation
Correct option: D
365 nm
Option D (365 nm) is correct because the most energetic photon corresponds to the shortest wavelength in the Balmer series, which occurs at the series limit (n = infinity).
115 Medium
115 of 248
Understanding +4 | -1
Assertion: For the scattering of α\alpha-particles at large angles, only the nucleus of the atom is responsible.
Reason: Nucleus is very heavy in comparison to electrons.
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 nucleus is responsible for large-angle scattering due to its heavy mass compared to electrons.
116 Medium
116 of 248
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.
117 Medium
117 of 248
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.
118 Medium
118 of 248
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.
119 Medium
119 of 248
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}}}.
120 Medium
120 of 248
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.
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