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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 211-240 of 248 questions

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211 Medium
211 of 248
Application +4 | -1
If the wavelength of the first line of the Balmer series in the hydrogen spectrum is λ,\lambda , then the wavelength of the first line of the Lyman series is
View answer and explanation
Correct option: B
(5/27)λ\left( {5/27} \right)\lambda
Option B ((5/27)λ\left( {5/27} \right)\lambda) is correct because it is calculated by finding the Rydberg constant from the Balmer series and then using it to find the wavelength of the Lyman series.
212 Medium
212 of 248
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 correct because Moseley's law states that the square root of the frequency of characteristic X-rays is proportional to the atomic number of the element.
213 Medium
213 of 248
Application +4 | -1
The electron beam in a colour TV is accelerated through 32 kV32 \mathrm{~kV} and then strikes the screen. What is the wavelength of the most energetic X-ray photon?
View answer and explanation
Correct option: A
38.8 pm38.8 \mathrm{~pm}
The correct answer is 38.8 pm. The wavelength of the most energetic X-ray photon is calculated using the energy gained by the electron beam.
214 Medium
214 of 248
Application +4 | -1
Calculate the shortest wavelength of Lyman series. Given Rydberg's constant, R=10967700 m−1R=10967700 \mathrm{~m}^{-1}.
View answer and explanation
Correct option: A
911.6A∘911.6 \mathop {\text{A}}\limits^ \circ
Option A (911.6A∘911.6 \mathop {\text{A}}\limits^ \circ) is correct because it is calculated using the Rydberg formula for the shortest wavelength in the Lyman series (n = infinity).
215 Medium
215 of 248
Understanding +4 | -1
As an electron makes a transition from an excited state to the ground state of hydrogen like atom/ion
View answer and explanation
Correct option: A
Its kinetic energy increases but potential energy and total energy decrease
Option A (Its kinetic energy increases but potential energy and total energy decrease) is correct because as the electron transitions to a lower energy level, its kinetic energy increases, while potential and total energy decrease.
216 Medium
216 of 248
Understanding +4 | -1
The graph between priciple quantum number and radius of orbit is :
View answer and explanation
Correct option: C
Parabolic curve
Option C is correct because the radius of an orbit is proportional to the square of the principal quantum number, resulting in a parabolic relationship.
217 Medium
217 of 248
Application +4 | -1
The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6×10−16s1.6 \times {10^{ - 16}}\mathrm{s} The frequency of revolution of the electron in its first excited state (in Hz) is
View answer and explanation
Correct option: D
7.8×10147.8 \times {10^{14}}
Option D (7.8×10147.8 \times {10^{14}}) is correct because it's the frequency of revolution in the first excited state, calculated using the relationship between time period and orbit number.
218 Medium
218 of 248
Application +4 | -1
A particle of mass mm moves in a circular orbit in a central potential field U(r)=U0r4U(r)=U_0 r^4. If Bohr's quantization conditions are applied, radii of possible orbitals rnr_n vary with n1/αn^{1 / \alpha}, where α\alpha is ......
View answer and explanation
Correct option: A
3
Option A (3) is correct because by applying Bohr's quantization condition and the given potential, the radius rr is proportional to n1/3n^{1/3}.
219 Medium
219 of 248
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 energy, and thus the penetrating power, of X-rays is directly proportional to the accelerating voltage.
220 Medium
220 of 248
Application +4 | -1
An electron beam of energy 10 KeV10 \mathrm{~KeV} is incident on metallic foil. If the interatomic distance is 0.55 A∘0.55 ~\mathop {\text{A}}\limits^ \circ. Find the angle of diffraction.
View answer and explanation
Correct option: B
sin⁡−1(0.22)\sin ^{-1}(0.22)
Option B (sin⁡−1(0.22)\sin ^{-1}(0.22)) is correct because it correctly uses the de Broglie wavelength and Bragg's law to find the diffraction angle.
221 Medium
221 of 248
Understanding +4 | -1
On the basis of Bohr’s atomic model the radius of 3rd orbit of any atom is
View answer and explanation
Correct option: B
Nine times of radius of first orbit
Option B is correct because according to Bohr's model, the radius of the nth orbit is proportional to n2n^2, so the radius of the 3rd orbit is nine times the radius of the first orbit.
222 Medium
222 of 248
Understanding +4 | -1
Assertion (A): The positively charged nucleus of an atom has a radius of almost 10−15 m10^{-15} \mathrm{~m}.

Reason (R): In α\alpha - particle scattering experiment, the distance of closest approach for α\alpha-particles is 10−15 m10^{-15} \mathrm{~m}.
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 distance of closest approach in the alpha-particle scattering experiment provides an estimate of the nuclear radius.
223 Medium
223 of 248
Understanding +4 | -1
Which of the following pairs is an isobar
View answer and explanation
Correct option: B
1H2_1{H^2} and 1H3_1{H^3}
Option D (15P30_{15}{P^{30}} and 14Si30_{14}S{i^{30}}) is correct because isobars are nuclides with the same mass number (A) but different atomic numbers (Z).
224 Medium
224 of 248
Application +4 | -1
Calculate the period of revolution of an electron revolving in the first orbit of hydrogen atom. Given radius of first orbit =0.53A∘=0.53 \mathop {\text{A}}\limits^ \circ and c=3×108 ms−1c=3 \times 10^8 \mathrm{~ms}^{-1}.
View answer and explanation
Correct option: A
1.52×10−16 s1.52 \times 10^{-16} \mathrm{~s}
Option A (1.52×10−16 s1.52 \times 10^{-16} \mathrm{~s}) is correct because it is the period calculated using the formula T=2πrvT = \frac{2 \pi r}{v} with the given radius and calculated velocity.
225 Medium
225 of 248
Application +4 | -1
A stationary hydrogen atom emits photon corresponding to the first line of Lyman series. If RR is the Rydberg’s constant and MM is the mass of the atom, then the velocity acquired by the atom is (neglect energy absorbed by the photon)
View answer and explanation
Correct option: B
3Rh4M\frac{{3Rh}}{{4M}}
Option B (3Rh4M\frac{{3Rh}}{{4M}}) is correct because it applies conservation of momentum and the Rydberg formula to find the recoil velocity of the hydrogen atom.
226 Medium
226 of 248
Application +4 | -1
Find the wavelengths present in the radiation emitted
when hydrogen atoms excited to n=3n = 3 states return to
their ground states.
View answer and explanation
Correct option: A
103 nm, 122 nm, 654 nm
Option A (103 nm, 122 nm, 654 nm) is correct because these wavelengths correspond to the transitions from n=3 to n=1, n=3 to n=2, and n=2 to n=1 in the hydrogen atom.
227 Medium
227 of 248
Application +4 | -1
Hydrogen atom emits blue light when it changes from n=4n = 4 energy level to the n=2n = 2 level. Which colour of light would the atom emit when it changes from the n=5n = 5 level to the n=2n = 2 level
View answer and explanation
Correct option: D
Violet
Option D (Violet) is correct because transitions to lower energy levels release more energy, and the transition from n=5 to n=2 releases more energy than the transition from n=4 to n=2, resulting in a higher frequency (violet) light.
228 Medium
228 of 248
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 is correct. The de-Broglie wavelength equation can be expressed as λ = 1.227/√V nm, where V is the accelerating potential in volts.
229 Medium
229 of 248
Application +4 | -1
Taking the Bohr radius as a0=53 pma_0=53\, {pm}, the radius of Li++{Li}^{++}ion in its ground state, on the basis of Bohr's model, will be about
View answer and explanation
Correct option: C
18 pm18 \,{pm}
Option C (18 pm18 \,{pm}) is correct because the radius of the Li++{Li}^{++} ion is calculated using Bohr's model formula, considering the atomic number and the ground state.
230 Medium
230 of 248
Remembering +4 | -1
The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?
View answer and explanation
Correct option: D
Balmer series
Option D (Balmer series) is correct because the Balmer series corresponds to transitions to the n=2 energy level, resulting in photons in the visible region of the electromagnetic spectrum.
231 Medium
231 of 248
Application +4 | -1
X-rays of wavelength 0.140 nm are scattered from a block of carbon. What will be the wavelenghts of X-rays scattered at 900{90^0}?
View answer and explanation
Correct option: B
0.142 nm
Option B (0.142 nm) is correct because it calculates the wavelength shift due to Compton scattering at 90 degrees.
232 Medium
232 of 248
Application +4 | -1
Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number nn changes by 2.

(a) Find the smallest wavelength emitted by hydrogen.

(b) List the wavelengths emitted by hydrogen in the visible range (380 nm(380 \mathrm{~nm} to 780 nm780 \mathrm{~nm}).
View answer and explanation
Correct option: A
(a) 103 nm;103 \mathrm{~nm}; (b) 487 nm487 \mathrm{~nm}
Option A ((a) 103 nm;103 \mathrm{~nm}; (b) 487 nm487 \mathrm{~nm}) is correct because the smallest wavelength and the wavelength in the visible range are calculated correctly based on the given transition rule.
233 Medium
233 of 248
Understanding +4 | -1
In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by
View answer and explanation
Correct option: A
increasing the potential difference between the anode and filament
Option A is correct because increasing the potential difference between the anode and filament increases the kinetic energy, and thus the velocity, of the electrons.
234 Medium
234 of 248
Understanding +4 | -1
If impact parameter is zero, then angle of deviation in α\alpha-particle scattering experiment, is
View answer and explanation
Correct option: D
180∘180^\circ
Option D (180∘180^\circ) is correct because if the impact parameter is zero, the alpha particle will retrace its path, resulting in a deviation of 180 degrees.
235 Medium
235 of 248
Application +4 | -1
A particular hydrogen-like ion emits radiation of
frequency 2.467×1015 Hz2.467 \times 10^{15} \text{ Hz} when it makes transition from
n=2n = 2 to n=1.n = 1. What will be the frequency of the radiation
emitted in a transition from n=3n = 3 to n=1?n = 1 ?
View answer and explanation
Correct option: A
2.92×1015  Hz2.92 \times {10^{15}}\,\,{\text{Hz}}
Option A (2.92×1015 Hz2.92 \times {10^{15}}\,{\text{Hz}}) is correct because it is calculated using the Rydberg formula for hydrogen-like ions, relating frequency to the difference in inverse squares of principal quantum numbers.
236 Medium
236 of 248
Application +4 | -1
Which element has a KαK_\alpha x-ray line whose wavelength is 0.180 nm0.180 \mathrm{~nm} ?
View answer and explanation
Correct option: A
cobalt
Option A (cobalt) is correct because its atomic number (Z=27) satisfies the approximate Moseley's law for K-alpha X-ray wavelength.
237 Medium
237 of 248
Remembering +4 | -1
Intensity of XX-rays depends upon the number of
View answer and explanation
Correct option: A
Electrons
Option A (Electrons) is correct because the intensity of X-rays is directly related to the number of electrons striking the target material.
238 Medium
238 of 248
Remembering +4 | -1
The value of series limit in the case of paschen series is:
View answer and explanation
Correct option: C
822 nm
Option C (822 nm) is correct because the series limit for the Paschen series is calculated using the Rydberg formula with n1=3 and n2=infinity.
239 Medium
239 of 248
Application +4 | -1
If The uncertainity in the position of an electron is 10−10 m10^{-10} {~m} then the uncertainity in its momentum (in kgms−1{kgms}^{-1} ) measurement will be nearly.
View answer and explanation
Correct option: B
10−2410^{-24}
Option B (10−2410^{-24}) is correct because it is calculated using the Heisenberg uncertainty principle.
240 Medium
240 of 248
Understanding +4 | -1
The energy of a photon of characteristic XX-rays from a Coolidge tube comes from
View answer and explanation
Correct option: D
An electronic transition of the target atom
Option D (An electronic transition of the target atom) is correct because characteristic X-rays are produced when electrons transition between energy levels in the target atom.
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