Skip to main content
Back to STRUCTURE OF ATOM

Your Progress

0 / 297 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 61-90 of 297 questions

Preparing practice and saved progress…

61 Medium
61 of 297
Application +4 | -1
What will be the ratio of the wavelength of the first lineto that of the second line of Paschen series of H{H} atom?
View answer and explanation
Correct option: A
256:175256: 175
Option A (256:175) is correct because it is the correct ratio of the wavelengths calculated using the Rydberg formula for the Paschen series.
62 Medium
62 of 297
Remembering +4 | -1
With respect to atomic spectrum, each line in the Lyman series is due to electrons returning
View answer and explanation
Correct option: C
From a particular higher energy level to n=1n=1
Option C (From a particular higher energy level to n=1n=1) is correct because the Lyman series corresponds to electronic transitions from higher energy levels to the n=1 energy level.
63 Medium
63 of 297
Analyzing +4 | -1
The decreasing order (highest first) from the values of e/m{e} / {m} (charge/mass) for electron (e),(e), proton (p)(p), neutron (n)(n) and alpha particle (α)(\alpha) is
View answer and explanation
Correct option: A
e,p,α,ne, p, \alpha, n
Option A is correct because the charge-to-mass ratio (e/m) decreases in the order: electron > proton > alpha particle > neutron.
64 Medium
64 of 297
Application +4 | -1
Ratio of the radius of first orbit of Li2+\mathrm{Li}^{2+} to the third orbit of He+\mathrm{He}^{+}ions will be
View answer and explanation
Correct option: D
2:272: 27
Option D (2:27) is correct because it is calculated by using the formula for the radius of Bohr orbits and taking the ratio.
65 Medium
65 of 297
Application +4 | -1
Calculate the energy of Li+2{Li}^{+2} atom for 2nd2^{\text {nd}} excited state.
View answer and explanation
Correct option: A
−13.6 eV-13.6 \mathrm{~eV}
Option A (-13.6 eV) is correct because it is calculated using the formula for energy levels in hydrogen-like atoms, considering the atomic number and the excited state.
66 Medium
66 of 297
Understanding +4 | -1
The correct set of quantum numbers for 4d4 d-electrons are
View answer and explanation
Correct option: D
4,2,1,−1/24,2,1,-1 / 2
Option D (4,2,1,−1/24,2,1,-1 / 2) is correct because for a 4d electron, n=4, l=2, m can be -2, -1, 0, 1, 2, and s can be +1/2 or -1/2.
67 Medium
67 of 297
Analyzing +4 | -1
Energies of 2S orbital in IA group is
View answer and explanation
Correct option: A
E2s(H)>E2s(Li)>E2s(Na)>E2s(K){E_{2s}}\left( H \right) > {E_{2s}}\left( {Li} \right) > {E_{2s}}\left( {Na} \right) > {E_{2s}}\left( K \right)
Option A is correct because as the atomic number increases down the group, the energy of the 2s orbital decreases due to increased nuclear charge and shielding.
68 Medium
68 of 297
Remembering +4 | -1
Compared to mass of lightest nucleus the mass of an electron is only:
View answer and explanation
Correct option: C
1/18001/1800
Option C (1/1800) is correct because the mass of an electron is approximately 1/1836 the mass of a hydrogen atom (lightest nucleus).
69 Medium
69 of 297
Understanding +4 | -1
The total number of angular nodes and radial nodes present in 3p orbital
View answer and explanation
Correct option: B
2
Option B (2) is correct because a 3p orbital has one angular node (l=1) and one radial node (n-l-1 = 3-1-1 = 1).
70 Medium
70 of 297
Understanding +4 | -1
The fraction of volume occupied by the nucleus with respect to the total volume of an atom is
View answer and explanation
Correct option: B
10−15{10^{ - 15}}
Option B is correct because the nucleus is extremely small compared to the overall size of the atom, occupying a fraction of approximately 10-15.
71 Medium
71 of 297
Application +4 | -1
The de-Broglie wavelength of the electron in the ground
state of hydrogen atom is (KE = 13.6 eV)
View answer and explanation
Correct option: C
0.3329 nm
Option C (0.3329 nm) is correct because it accurately calculates the de Broglie wavelength using the given kinetic energy of the electron.
72 Medium
72 of 297
Analyzing +4 | -1
A sample of hydrogen atoms containing all the atoms in a particular excited state absorb radiations of a particular wavelength by which the atoms get excited to another excited state. When the atoms finally de-excite to the ground state, they emit the radiations of 10 different wavelengths. Out of these 10 radiations, 7 have wavelengths shorter than the absorbed radiation and 2 have wavelength longer than the absorbed radiation. The orbit number for the initial excited state of atoms is
View answer and explanation
Correct option: A
3
The correct answer is A (3) because the atom absorbs radiation to reach the 5th orbit and emits 2 wavelengths longer than the absorbed one, indicating the initial state was the 3rd orbit.
73 Medium
73 of 297
Application +4 | -1
The wave number of first line of Balmer series of Hydrogen is 15200  cm−115200\,\,c{m^{ - 1}}. The wavenumber of the first line of 15200 cm−115200\,c{m^{ - 1}} ion is
View answer and explanation
Correct option: B
60800 cm−160800\,c{m^{ - 1}}
Option B (60800 cm−160800\,c{m^{ - 1}}) is correct because the wavenumber is proportional to Z2Z^2, where Z is the atomic number.
74 Medium
74 of 297
Application +4 | -1
A dust particle has mass equal to 10−11 g10^{-11} \mathrm{~g}, diameter 10−4 cm10^{-4} \mathrm{~cm} and velocity 10−4 cm s−110^{-4} \mathrm{~cm} \mathrm{~s}^{-1}. The error in measurement of velocity is 0.1%0.1 \%. What will be the uncertainty in its position?
View answer and explanation
Correct option: D
0.527×10−9 cm0.527 \times 10^{-9} \mathrm{~cm}
Answer D (0.527 x 10-9 cm) is correct because it is calculated using the Heisenberg uncertainty principle, considering the error in velocity measurement.
75 Medium
75 of 297
Analyzing +4 | -1
Evaluate the following statements related to group 14 elements for their correctness.

(A)Covalent radius decreases down the group from C\mathrm{C} to Pb\mathrm{Pb} in a regular manner.

(B) Electronegativity decreases from C\mathrm{C} to Pb\mathrm{Pb} down the group gradually.

(C) Maximum covalence of C\mathrm{C} is 4 whereas other elements can expand their covalence due to presence of dd orbitals.

(D) Heavier elements do not form pπ−pπ\mathrm{p} \pi-\mathrm{p} \pi bonds.

(E) Carbon can exhibit negative oxidation states.
Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(C), (D) and (E) Only
Option A ((C), (D) and (E) Only) is correct because statements C, D, and E are accurate descriptions of group 14 elements' properties.
76 Medium
76 of 297
Understanding +4 | -1
Which one of the following is not the characteristic of Planck's quantum theory of radiation?
View answer and explanation
Correct option: A
The energy is not absorbed or emitted in whole number or multiple of quantum.
Option A is correct because Planck's quantum theory states that energy is absorbed or emitted in whole number multiples of a quantum, not just any amount.
77 Medium
77 of 297
Application +4 | -1
In the ground state of Cu+\mathrm{C}{\mathrm{u}^ + }, the number of shell occupied, sub-shells occupied, fillied orbitals and unpaired electrons respectively are
View answer and explanation
Correct option: C
3,6,14,0
Option C (3,6,14,0) is correct because Cu+ has 3 shells, 6 subshells, 14 filled orbitals and 0 unpaired electrons.
78 Medium
78 of 297
Analyzing +4 | -1
Statement (S): Wavelength of I line of Humphrey series is more than I line of Lyman series in H-atom.

Explanation (E): ΔE=hcλ.\Delta E=\frac{h c}{\lambda}.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation of S
Option C (Both S and E are correct and E is correct explanation of S) is correct because the energy difference in the Humphrey series is smaller than in the Lyman series, leading to a longer wavelength, and the relationship between energy and wavelength is given by ΔE = hc/λ.
79 Medium
79 of 297
Application +4 | -1
How many moles of electrons weigh one milligram?
View answer and explanation
Correct option: B
1029.1×6.02\frac{10^2}{9.1 \times 6.02}
Option B is correct because it accurately calculates the number of moles of electrons required to weigh one milligram based on the mass of a single electron and Avogadro's number.
80 Medium
80 of 297
Understanding +4 | -1
Which shell would be the first to have ' gg ' sub-shell?
View answer and explanation
Correct option: D
O\mathrm{O}
Option D (O) is correct because the 'g' subshell corresponds to l=4, and the first shell to have l=4 is the n=5 (O) shell.
81 Medium
81 of 297
Analyzing +4 | -1
Among the following transitions in hydrogen and
hydrogen-like ion spectrum, which one emits light
of longest wavelength ?
View answer and explanation
Correct option: C
n=4\mathrm{n}=4 to n=3\mathrm{n}=3 for He+\mathrm{He}^{+}
The transition with the smallest energy difference will emit light of the longest wavelength. Among the given options, n=4 to n=3 for He+ has the smallest energy difference considering the Z2 factor, making option C the correct answer.
82 Medium
82 of 297
Analyzing +4 | -1
Of the following transitions in hydrogen atom, the one which gives an absorption line of lowest frequency is :
View answer and explanation
Correct option: B
n=3  to  n=8n = 3\;\mathrm{to\;}n = 8
The correct answer is B (n=3  to  n=8n = 3\;\mathrm{to\;}n = 8) because the energy difference between these levels is the smallest among the absorption transitions, resulting in the lowest frequency absorption line.
83 Medium
83 of 297
Remembering +4 | -1
The radiation having maximum wavelength is
View answer and explanation
Correct option: B
Radio waves
Option B (Radio waves) is correct because radio waves have the longest wavelength compared to UV rays, X-rays, and IR rays.
84 Medium
84 of 297
Application +4 | -1
The uncertainty in the position of an electron (mass == 9.1×10−28 g9.1 \times 10^{-28} {~g} ) moving with a velocity of 3.0×104 cm s−13.0 \times 10^4 {~cm} {~s}^{-1} accurate upto 0.011%0.011 \% will be
View answer and explanation
Correct option: C
0.175 cm0.175 {~cm}
Option C (0.175 cm0.175 {~cm}) is correct because it is calculated using the Heisenberg uncertainty principle, relating uncertainty in position and momentum.
85 Medium
85 of 297
Application +4 | -1
The electronic velocity in the fourth Bohr's orbit of hydrogen is v. The velocity of the electron in the first orbit would be:
View answer and explanation
Correct option: A
4v
The correct answer is 4v because the electronic velocity in a Bohr orbit is inversely proportional to the principal quantum number n.
86 Medium
86 of 297
Application +4 | -1
Ratio of radii of second and first Bohr orbits of H – atom
View answer and explanation
Correct option: B
4
Option B (4) is correct because the radius of the nth Bohr orbit is proportional to n², so the ratio of the radii of the second and first Bohr orbits is 2²/1² = 4.
87 Medium
87 of 297
Application +4 | -1
Arrange α−\alpha - particle (α)(\alpha ), ,electron (e−),({e^ - }), proton (p) and neutron (n) in increasing order of their e/m value
View answer and explanation
Correct option: B
n<α<p<e−n < \alpha < p < {e^ - }
The correct answer is B (n<α<p<e−n < \alpha < p < {e^ - }). The e/m ratio depends on the charge and mass of the particle.
88 Medium
88 of 297
Understanding +4 | -1
Choose the correct energy order for 2s2 s orbital
View answer and explanation
Correct option: A
He(2s)>Li(2s)>Be(2s){He}(2 s)>{Li}(2 s)>{Be}(2 s)
Option A (He(2s)>Li(2s)>Be(2s){He}(2 s)>{Li}(2 s)>{Be}(2 s)) is correct because the energy of an orbital decreases with increasing nuclear charge for the same principal quantum number.
89 Medium
89 of 297
Application +4 | -1
What will be the wave length of an electron moving with 110\frac{1}{10} th of velocity of light?
View answer and explanation
Correct option: A
2.43×10−11 m2.43 \times 10^{-11} \mathrm{~m}
Option A (2.43×10−11 m2.43 \times 10^{-11} \mathrm{~m}) is correct because the wavelength is calculated using the de Broglie wavelength formula, λ=h/mv\lambda = h/mv, with the given velocity.
90 Medium
90 of 297
Remembering +4 | -1
According to Bohr's theory, the electronic energy of H\mathrm{H}-atom in Bohr's orbit is given by
View answer and explanation
Correct option: B
En=−2.179×10−18×Z2n2J\mathrm{E_n}=-\frac{2.179 \times 10^{-18} \times \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{J}
Option B (En=−2.179×10−18×Z2n2J\mathrm{E_n}=-\frac{2.179 \times 10^{-18} \times \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{J}) is correct because it represents the correct formula for electronic energy of a hydrogen atom in Bohr's orbit.
Revision notes and key takeaways

Review STRUCTURE OF ATOM in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 297 questions

Your practice

0 / 297 answered · 297 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.