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NEET • PHYSICS • ATOMS

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91 Medium
91 of 130
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 Moseley's law states that the frequency of characteristic X-rays is proportional to the square of the atomic number minus a constant.
92 Medium
92 of 130
Remembering +4 | -1
In the lowest energy level of hydrogen atom, the electron has the angular momentum
View answer and explanation
Correct option: C
h/2πh/2\pi
Option C (h/2πh/2\pi) is correct because the angular momentum in the lowest energy level (n=1) is quantized and equal to h/2πh/2\pi.
93 Medium
93 of 130
Analyzing +4 | -1
Assertion (A): According to classical theory, the proposed path of an electron in Rutherford atom model will be parabolic.

Reason (R): According to electromagnetic theory an accelerated particle continuously emits radiation.
View answer and explanation
Correct option: B
Both (A) and (R) are true and (R) is not the correct explanation of (A)
Option B (Both (A) and (R) are true and (R) is not the correct explanation of (A)) is correct because while both the parabolic path and radiation emission are true statements, the radiation emission doesn't directly cause the parabolic path.
94 Medium
94 of 130
Application +4 | -1
The electric potential between a proton and an electron is given by V=V0  ln⁡  rr0V = {V_0}\,\,\ln \,\,\frac{r}{{{r_0}}} , where r0{r_0} is a constant. Assuming Bohr’s model to be applicable, write variation of rn{r_n} with n,nn,n being the principal quantum number
View answer and explanation
Correct option: A
rn  ∝  n{r_n}\,\, \propto \,\,n
Option A (rn  ∝  n{r_n}\,\, \propto \,\,n) is correct because, according to Bohr's model and the given potential, the radius is directly proportional to the principal quantum number.
95 Medium
95 of 130
Understanding +4 | -1
Penetrating power of X-rays depends on
View answer and explanation
Correct option: B
Applied potential difference
Option B (Applied potential difference) is correct because the penetrating power of X-rays is directly related to their energy, which is determined by the applied potential difference.
96 Medium
96 of 130
Remembering +4 | -1
The value of structure constant is
View answer and explanation
Correct option: A
1137\frac{1}{137}
Option A (1137\frac{1}{137}) is correct because it is the approximate value of the fine-structure constant, a fundamental physical constant.
97 Medium
97 of 130
Analyzing +4 | -1
Electrons in a certain energy level n=n1n = {n_1}, can emit 3 spectral lines\mathrm{3}\,\mathrm{ spectral}\,\mathrm{ lines}. When they are in another energy level, n=n2n = {n_2}. They can emit 6 spectral lines\mathrm{6}\,\mathrm{ spectral}\,\mathrm{ lines}. The orbital speed of the electrons in the two orbits are in the ratio
View answer and explanation
Correct option: A
4:3\mathrm{4 : 3}
Option A (4 : 3) is correct because the ratio of orbital speeds is inversely proportional to the ratio of the principal quantum numbers, which can be found from the number of spectral lines emitted.
98 Medium
98 of 130
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 the wavelength of the 2nd line of the Lyman series is related to the Rydberg constant, and the wavelength of the last line of the Balmer series is also related to the Rydberg constant. By setting up the equations and solving for the last line of the Balmer series in terms of λ\lambda, the correct answer is obtained.
99 Medium
99 of 130
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 C (1λβ=1λα+1λ′α\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}) is correct because the energy difference between levels corresponds to the sum of energies of individual transitions.
100 Medium
100 of 130
Understanding +4 | -1
Whenever a hydrogen atom emits a photon in the Balmer series
View answer and explanation
Correct option: C
It must emit another photon in the Lyman series
Option C (It must emit another photon in the Lyman series) is correct because after emitting a Balmer series photon, the electron must eventually transition to the ground state, emitting Lyman series photons.
101 Hard
101 of 130
Remembering +4 | -1
The ionization energy for second He electron is:
View answer and explanation
Correct option: C
54.4 eV
Option C (54.4 eV) is correct because the ionization energy for the second He electron is the same as the ionization energy of hydrogen with Z=2, which is 13.6eV∗Z2=13.6eV∗4=54.4eV13.6 eV * Z^2 = 13.6 eV * 4 = 54.4 eV.
102 Easy
102 of 130
Application +4 | -1
An XX-ray tube is run at 50 kV50 \mathrm{~kV}. The current flowing in it is 20 mA20 \mathrm{~mA}. The power of the tube is:
View answer and explanation
Correct option: A
1000 W1000 \mathrm{~W}
Option A (1000 W) is correct because the power of the X-ray tube is the product of the voltage and the current, which is 50 kV * 20 mA = 1000 W.
103 Medium
103 of 130
Understanding +4 | -1
When a hydrogen atom is raised from the ground state to an excited state
View answer and explanation
Correct option: A
P.E.\mathrm{P}\mathrm{.E}\mathrm{.} increases and K.E.\mathrm{K}\mathrm{.E}\mathrm{.} decreases
Option A (P.E.\mathrm{P}\mathrm{.E}\mathrm{.} increases and K.E.\mathrm{K}\mathrm{.E}\mathrm{.} decreases) is correct because as the electron moves to a higher energy level, its potential energy becomes less negative (increases) and its kinetic energy decreases.
104 Medium
104 of 130
Understanding +4 | -1
An electron in the n=1n = 1 orbit of hydrogen atom is bounded by 13.6  eV13.6\,\,eV. Energy requires to ionize it is
View answer and explanation
Correct option: A
13.6  eV13.6\,\,eV
Option A (13.6  eV13.6\,\,eV) is correct because the ionization energy is the energy required to remove an electron from the atom, which is equal to the binding energy in the ground state.
105 Medium
105 of 130
Understanding +4 | -1
In Bohr’s model of hydrogen atom, let PEPE represents potential energy and TETE the total energy. In going to a higher level
View answer and explanation
Correct option: B
PEPE increases, TETE increases
Option B (PEPE increases, TETE increases) is correct because as the electron goes to a higher orbit, its potential energy becomes less negative (increases) and its total energy also increases (becomes less negative).
106 Medium
106 of 130
Analyzing +4 | -1
The ionization energy of the electron in the hydrogen atom in its ground state is 13.6  eV13.6\,\,eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths\mathrm{6 }\,\mathrm{wavelengths}. Maximum wavelength of emitted radiation corresponds to the transition between
View answer and explanation
Correct option: D
n=4n = 4 to n=3n = 3 states
Option D (n=4n = 4 to n=3n = 3 states) is correct because the transition with the smallest energy difference will have the maximum wavelength, and among the given options, the transition from n=4 to n=3 has the smallest energy difference.
107 Medium
107 of 130
Remembering +4 | -1
In Bohr’s model of hydrogen atom, which of the following represents the variation of nth orbit radius with n?
View answer and explanation
Correct option: D
rn∝n2{r_n} \propto {n^2}
Option D (rn∝n2{r_n} \propto {n^2}) is correct because the radius of the nth orbit in Bohr's model is proportional to the square of the principal quantum number n.
108 Medium
108 of 130
Application +4 | -1
The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:
View answer and explanation
Correct option: B
79 eV
Option C (109 eV) is correct because the total energy required is the sum of the energy to remove the first electron and 2.2 times the energy to remove the first electron, which equals 3.2 times the energy to remove the first electron. Given the energy to remove the first electron is 24.6 eV, the total energy is approximately 79 eV.
109 Medium
109 of 130
Understanding +4 | -1
Which of the following is true
View answer and explanation
Correct option: B
Paschen series is a line spectrum in the infrared
Option B (Paschen series is a line spectrum in the infrared) is correct because the Paschen series involves transitions to the n=3 energy level, which results in photons with energies in the infrared region.
110 Medium
110 of 130
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.
111 Easy
111 of 130
Application +4 | -1
XX-rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength 22.85 nm22.85 \mathrm{~nm} attributed to the KαK_\alpha line of an impurity element. The KαK_\alpha line's of copper (Z=29)(Z=29) and molybdenum (Z=42)(Z=42) have wavelengths 15.42 nm15.42 \mathrm{~nm} and 7.12 nm7.12 \mathrm{~nm} respectively. The atomic number of the impurity element is:
View answer and explanation
Correct option: A
24
Option A (24) is correct because the atomic number of the impurity element can be found using Moseley's law, and solving for Z yields approximately 24.
112 Medium
112 of 130
Analyzing +4 | -1
The equation λ=1.227x nm\lambda=\frac{1.227}{x} \mathrm{~nm} can be used to find the de- Broglie wavelength of an electron.
In this equation xx stands for:
Where m = mass of electron
P = momentum of electron
K = Kinetic energy of electron
V = Accelerating potential in volts for electron
View answer and explanation
Correct option: D
V\sqrt{V}
Option D (V\sqrt{V}) is correct because the de-Broglie wavelength of an electron is inversely proportional to the square root of the accelerating potential (V).
113 Medium
113 of 130
Remembering +4 | -1
According to Moseley’s law of XX-rays the frequency (v)\left( v \right) of particular characteristic XX-ray and the atomic number (Z)\left( Z \right) of the element depend on each other as
View answer and explanation
Correct option: D
v=kZ\sqrt v = kZ
Option D (v=kZ\sqrt v = kZ) is the correct representation of Moseley's law, which states that the square root of the frequency of X-rays is proportional to the atomic number.
114 Easy
114 of 130
Application +4 | -1
A particle AA has a charge qq and particle BB has charge 4q4 q with each of them having the mass m{m}. When they are allowed to fall from rest through the same potential difference, the ratio of their speeds vA:vBv_A: v_B will be
View answer and explanation
Correct option: C
1:21: 2
Option C (1: 2) is correct because the kinetic energy gained by a charged particle falling through a potential difference is proportional to the charge, and the speed is proportional to the square root of the kinetic energy divided by the mass.
115 Medium
115 of 130
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. The longest wavelength corresponds to the transition from n=2 to n=1, and the shortest wavelength corresponds to the transition from n=∞\infty to n=1.
116 Medium
116 of 130
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 D (36R\frac{{36}}{R}) is correct because the maximum wavelength in the Balmer series corresponds to the transition from n=3 to n=2. Using the Rydberg formula, 1λ=R(122−132)=R(14−19)=R(536)\frac{1}{\lambda} = R(\frac{1}{2^2} - \frac{1}{3^2}) = R(\frac{1}{4} - \frac{1}{9}) = R(\frac{5}{36}). Therefore, λ=365R\lambda = \frac{36}{5R}.
117 Easy
117 of 130
Application +4 | -1
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th 5^{\text {th }} excited state of a hydrogen atom is :
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because the kinetic energy is the negative of the total energy, and the potential energy is twice the total energy, so the ratio is 1/2.
118 Medium
118 of 130
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
Option B (Both A and R are correct and R is the correct explanation of A) is correct because the wave nature of electrons, which allows them to exhibit interference and diffraction, was experimentally verified by the Davisson-Germer experiment.
119 Easy
119 of 130
Understanding +4 | -1
Cathode ray tube is operating at 5 KV5\,KV. Then the K.E. acquired by the electron is
View answer and explanation
Correct option: C
5 KeV
Option C (5 KeV) is correct because the kinetic energy acquired by an electron accelerated through a potential difference of 5 kV is 5 keV.
120 Easy
120 of 130
Understanding +4 | -1
Statement I: Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Statement II: According to classical physics all moving electrons radiate electromagnetic radiation.
View answer and explanation
Correct option: B
Statement I is true; Statement II is true; Statement II is not the correct explanation for Statement I
Option B (Statement I is true; Statement II is true; Statement II is not the correct explanation for Statement I) is correct because both statements are factually correct, but Statement II doesn't explain why Bohr postulated Statement I.
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