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JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom 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 211-240 of 297 questions

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211 Medium
211 of 297
Application +4 | -1
An electron is accelerated from a very low velocity (∼\sim zero speed) by the application of a potential difference of V\mathrm{V} volts. If the de Broglie wavelength should change (i.e., decrease) by 1.0%1.0 \% what percent increase in V\mathrm{V} causes it-
View answer and explanation
Correct option: D
2%2 \%
Option D (2%) is correct because the de Broglie wavelength is inversely proportional to the square root of the potential difference.
212 Medium
212 of 297
Remembering +4 | -1
Which experiment is responsible for finding out the charge on an electron?
View answer and explanation
Correct option: A
Millikan’s oil drop experiment
Option A (Millikan’s oil drop experiment) is correct because Millikan's oil drop experiment was designed to measure the charge of a single electron.
213 Medium
213 of 297
Application +4 | -1
The energy of a photon is 3×10−123 \times {10^{ - 12}} ergs. What is its wavelength in nm? (h=6.62×10−27ergs,  c=3×1010cm/s)\left( {h = 6.62 \times {{10}^{ - 27}}\mathrm{ergs},\;c = 3 \times {{10}^{10}}\mathrm{cm}/\mathrm{s}} \right)
View answer and explanation
Correct option: A
662
The correct answer is A (662) because the wavelength is calculated using the energy of the photon and the Planck-Einstein relation, resulting in 662 nm.
214 Medium
214 of 297
Application +4 | -1
What is the maximum number of electrons in an atom that can have the following quantum numbers n=4,  m1=+1?n = 4,\;{m_1} = + 1?
View answer and explanation
Correct option: D
6
Option D (6) is correct because with n=4 and ml=+1, we are looking at the p-orbital, which can hold a maximum of 6 electrons.
215 Medium
215 of 297
Application +4 | -1
The radius of Bohr's first orbit in H-atom is 0.053 nm. The radius of second orbit in He+\mathrm{H}{\mathrm{e}^ + } would be:
View answer and explanation
Correct option: C
0.1060 nm
Option C (0.1060 nm) is correct because the radius of the nth Bohr orbit is proportional to n²/Z, where n is the principal quantum number and Z is the atomic number.
216 Medium
216 of 297
Understanding +4 | -1
Assertion: Dalton's Atomic Theory was later modified when the existence of subatomic particles was discovered.

Reason: The discovery of subatomic particles showed that atoms are not indivisible and can be further broken down into smaller particles.
View answer and explanation
Correct option: B
Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion
Option B (Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion) is correct because while both statements are true, the reason doesn't fully explain why Dalton's theory needed modification.
217 Medium
217 of 297
Remembering +4 | -1
The radius of the stationary state which is also called Bohr radius is given by the expression rn=n2a0\mathrm{r_n=n^2 a_0} where the value of a0\mathrm{a}_0 is
View answer and explanation
Correct option: A
52.9 pm52.9 \mathrm{~pm}
Option A (52.9 pm52.9 \mathrm{~pm}) is correct because the Bohr radius, denoted as a0a_0, has a value of 52.9 picometers.
218 Medium
218 of 297
Application +4 | -1
Which hydrogen like species will have same radius as that of Bohr orbit hydrogen atom?
View answer and explanation
Correct option: B
n=2,  Be3+n = 2,\mathrm{\;B}{\mathrm{e}^{3 + }}
Option B (n=2,  Be3+n = 2,\mathrm{\;B}{\mathrm{e}^{3 + }}) is correct because it satisfies the condition where the radius of the hydrogen-like species is equal to the Bohr radius of the hydrogen atom.
219 Medium
219 of 297
Remembering +4 | -1
The expression for Bohr's frequency rule is
View answer and explanation
Correct option: A
v=ΔEh=E2−E1hv=\frac{\Delta E}{h}=\frac{E_2-E_1}{h}
Option A (v=ΔEh=E2−E1hv=\frac{\Delta E}{h}=\frac{E_2-E_1}{h}) is correct because it represents Bohr's frequency rule, which relates the frequency of emitted or absorbed radiation to the energy difference between two energy levels.
220 Medium
220 of 297
Application +4 | -1
It the radius of first Bohr orbit be ao{a_\mathrm{o}}, then the radius of the third orbit would be
View answer and explanation
Correct option: C
9ao9{a_\mathrm{o}}
Option C (9ao9{a_\mathrm{o}}) is correct because the radius of the nth Bohr orbit is proportional to n squared.
221 Medium
221 of 297
Understanding +4 | -1
S-I: Wavelength of limiting line of lyman series is less than wavelength of limiting line of Balmer series.

S-II: Rydberg constant value is same for all elements.
View answer and explanation
Correct option: C
S-I is correct and S-II is incorrect
The correct answer is C because the limiting line of the Lyman series has a shorter wavelength than the Balmer series, and the Rydberg constant depends on the element.
222 Medium
222 of 297
Remembering +4 | -1
The maximum number of electrons accommodated in all 5f-orbitals
View answer and explanation
Correct option: C
14
Option C (14) is correct because each orbital can accommodate a maximum of 2 electrons, and there are 7 f-orbitals (5f), so 7 * 2 = 14 electrons.
223 Medium
223 of 297
Application +4 | -1
In H\mathrm{H}-atom, the wave number ratio is 108:7108: 7 is for
View answer and explanation
Correct option: C
first Lyman and first Paschen transition.
Option C (first Lyman and first Paschen transition) is correct because the given wave number ratio matches the ratio calculated for the first transitions of the Lyman and Paschen series in the hydrogen atom spectrum.
224 Medium
224 of 297
Analyzing +4 | -1
Assertion: Ratio of anode rays is different for different gases.
Reason: Proton is the fundamental particle present in gases.
View answer and explanation
Correct option: B
Both assertion and reason are true but reason is not the correct explanation of assertion
Option B (Both assertion and reason are true but reason is not the correct explanation of assertion) is correct because both statements are true, but the reason doesn't explain the assertion.
225 Medium
225 of 297
Application +4 | -1
Angular momentum of 2nd Bohr orbit of H – atom is xx . Then find out angular momentum in first excited state of Li2+L{i^{2 + }} is
View answer and explanation
Correct option: D
xx
The correct answer is D (xx) because the angular momentum depends on the principal quantum number (n) and is independent of the atomic number (Z). Both H in the 2nd orbit and Li²⁺ in the first excited state have n=2.
226 Medium
226 of 297
Understanding +4 | -1
If an electron has spin quantum number of +12+ \frac{1}{2} and a magnetic quantum number of -1, it cannot be represented in an
View answer and explanation
Correct option: A
s-orbital
Option A (s-orbital) is correct because an s-orbital has an angular momentum quantum number (l) of 0, and therefore a magnetic quantum number (m) of 0, so m = -1 is not possible for an s-orbital.
227 Medium
227 of 297
Application +4 | -1
If the mass of an electron is 9.11×10−31 kg,9.11 \times {10^{ - 31}}\,{\text{kg}}, and uncertainty in position is 10−8m,{10^{ - 8}}{\text{m}}, the uncertainty in velocity would be:
View answer and explanation
Correct option: B
5.8×103ms−15.8 \times {10^3}{\text{m}}{{\text{s}}^{ - 1}}
Option B (5.8×103 ms−15.8 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}) is correct because it is calculated using the Heisenberg uncertainty principle.
228 Medium
228 of 297
Understanding +4 | -1
For 19F−1,16O−2,20Ne,{ }^{19} {F}^{-1},{ }^{16} {O}^{-2}, { }^{20} {Ne}, choose the correct statement.
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all three species, 19F-1, 16O-2, and 20Ne, have 10 electrons each, making them isoelectronic.
229 Medium
229 of 297
Application +4 | -1
The mass of a photon with wavelength 3.6 Å  is
View answer and explanation
Correct option: C
6.135×10−33kg6.135 \times {10^{ - 33}}\mathrm{kg}
Option C (6.135×10−33kg6.135 \times {10^{ - 33}}\mathrm{kg}) is correct because it is the mass calculated using the energy-mass equivalence and the photon's wavelength.
230 Medium
230 of 297
Application +4 | -1
A dye emits 50%50 \% of the absorbed energy as fluorescence. If the number of quanta absorbed and emitted out is in the ratio 1:21: 2 and it absorbs the radiation of wavelength ' xx ' A∘\mathop {\text{A}}\limits^ \circ, then the wavelength of the emitted radiation will be
View answer and explanation
Correct option: C
4x A∘4 x\,\mathop {\text{A}}\limits^ \circ
Option C (4x A∘4 x\,\mathop {\text{A}}\limits^ \circ) is correct because the emitted wavelength is four times the absorbed wavelength, derived from the energy and quantum ratio relationship.
231 Medium
231 of 297
Understanding +4 | -1
Which are in the ascending order of wavelength?
View answer and explanation
Correct option: B
Lyman limit, Balmer limit, Paschen limit in the hydrogen spectrum
Option B is correct because the wavelengths of the Lyman, Balmer, and Paschen limits increase in that order, corresponding to the ultraviolet, visible, and infrared regions, respectively.
232 Medium
232 of 297
Application +4 | -1
Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.
View answer and explanation
Correct option: A
27419.25 cm−127419.25 \mathrm{~cm}^{-1}
Option A (27419.25 cm−127419.25 \mathrm{~cm}^{-1}) is correct because the shortest wavelength corresponds to the largest energy transition, which in the Balmer series is from infinity to n=2.
233 Medium
233 of 297
Application +4 | -1
Which of the followings sets of quantum numbers represents the highest energy of an atom?
View answer and explanation
Correct option: B
n=3,l=2,m=1,s=+12n = 3,l = 2,m = 1,s = + \frac{1}{2}
The correct answer is B (n=3,l=2,m=1,s=+12n = 3, l = 2, m = 1, s = + \frac{1}{2}) because it represents the highest energy level based on the n+ln + l rule.
234 Medium
234 of 297
Application +4 | -1
How many sets of four quantum number are possible for the electrons present in He2−?\mathrm{H}{\mathrm{e}^{2 - }}?
View answer and explanation
Correct option: A
4
Option A (4) is correct because He2−\mathrm{H}{\mathrm{e}^{2 - }} has four electrons, and each electron can have a unique set of four quantum numbers according to Pauli's exclusion principle.
235 Medium
235 of 297
Analyzing +4 | -1
Which of the following has maximum energy
View answer and explanation
Correct option: C
An electron moving with velocity of light
Option C (An electron moving with velocity of light) has maximum energy because the kinetic energy of an electron moving at the speed of light is significantly higher than the energy of a photon with a wavelength of 10 nm or a frequency of 10 Hz.
236 Medium
236 of 297
Application +4 | -1
The radius of hydrogen atom in the ground state is 0.53A00.53{A^0}. The radius of Li2+L{i^{2 + }} ion (atomic number =3) in a similar state is
View answer and explanation
Correct option: A
0.176 Å
Option A (0.176 Å) is correct because the radius of Li2+Li^{2+} is calculated using the formula rn=r0×n2Z{r_n} = \frac{{{r_0} \times {n^2}}}{Z}.
237 Medium
237 of 297
Remembering +4 | -1
The presence of charge particles in the atoms was first confirmed by
View answer and explanation
Correct option: C
Faraday
Option C (Faraday) is correct because Faraday's experiments on electrolysis provided the first concrete evidence of charged particles within atoms.
238 Medium
238 of 297
Application +4 | -1
Time period of a wave is 5×10−3s5 \times {10^{ - 3}}s, what is the frequency?
View answer and explanation
Correct option: B
2×102s−12 \times {10^2}{\mathrm{s}^{ - 1}}
Option B (2×102s−12 \times {10^2}{\mathrm{s}^{ - 1}}) is correct because frequency is the inverse of the time period.
239 Medium
239 of 297
Application +4 | -1
A photon of light of wavelength 6000 Å has energy EE. What will be the wavelength of photon of a light which has energy of photon 4E4 E ?
View answer and explanation
Correct option: A
1500 Å
Option A (1500 Å) is correct because energy and wavelength are inversely proportional, so if the energy is quadrupled, the wavelength is reduced to one-fourth.
240 Medium
240 of 297
Remembering +4 | -1
Select the incorrect option among the following
View answer and explanation
Correct option: A
e/m\mathrm{e} / \mathrm{m} ratio of proton is 1.758820×1011Cg−11.758820 \times 10^{11} \mathrm{C} \mathrm{g}^{-1}
Option A (e/m\mathrm{e} / \mathrm{m} ratio of proton is 1.758820×1011Cg−11.758820 \times 10^{11} \mathrm{C} \mathrm{g}^{-1}) is correct because the given value is the e/m ratio for an electron, not a proton. The e/m ratio for a proton is much smaller.
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