Skip to main content
Back to STRUCTURE OF ATOM

Your Progress

0 / 592 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • STRUCTURE OF ATOM

Structure of atom practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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 541-570 of 592 questions
541 Medium
541 of 592
Remembering +4 | -1
Which of the following statement is/are correct about Hertz (Hz)(\mathrm{Hz})?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given statements about Hertz (Hz) are accurate and well-established.
542 Easy
542 of 592
Analyzing +4 | -1
Statement (S): Specific charge of α\alpha-particles is twice to that of proton.

Explanation (E): Specific charge is given by e/m.e / m.
View answer and explanation
Correct option: B
S is wrong but E is correct
Option B (S is wrong but E is correct) is correct because the specific charge of an alpha particle is half that of a proton, and the explanation correctly defines specific charge.
543 Medium
543 of 592
Remembering +4 | -1
Cathode rays have same charge to mass ratio as
View answer and explanation
Correct option: A
β\beta-rays
Option A (β\beta-rays) is correct because cathode rays are streams of electrons, and beta particles are also electrons emitted from the nucleus.
544 Medium
544 of 592
Application +4 | -1
The ratio of wavelength values of series limit lines (n2=∞)(n_2 = \infty) of Balmer series and Paschen series are
View answer and explanation
Correct option: A
0.172916667
Option A (04:09) is correct because the ratio of the wavelengths is found by using the Rydberg formula for the series limit of the Balmer and Paschen series.
545 Medium
545 of 592
Application +4 | -1
The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The possible energy value of electron in the excited state of Li2+\mathrm{L}{\mathrm{i}^{2 + }} is
View answer and explanation
Correct option: C
−30.6 eV- 30.6\,eV
Option C (-30.6 eV) is correct because it's calculated using the formula for energy levels in hydrogen-like atoms, considering the excited state (n=2) and the atomic number of Li2+ (Z=3).
546 Medium
546 of 592
Application +4 | -1
Electromagnetic radiations of wavelength 240 nm\mathrm{nm} are just sufficient to ionize sodium atom. The ionization energy of sodium (in kJ/mol\mathrm{kJ} / \mathrm{mol} ) is
View answer and explanation
Correct option: B
498.58
Option B (498.58) is correct because the ionization energy is calculated using the given wavelength and the formula I.E. = hc/λ.
547 Easy
547 of 592
Application +4 | -1
Which of the following particles has more electrons than neutrons
View answer and explanation
Correct option: C
O−2{O}^{-2}
Option C (O−2{O}^{-2}) is correct because it has 10 electrons and 8 neutrons, thus having more electrons than neutrons.
548 Easy
548 of 592
Understanding +4 | -1
The H-spectrum show
View answer and explanation
Correct option: D
Presence of quantised energy level
Option D (Presence of quantised energy level) is correct because the hydrogen spectrum provides evidence for the existence of discrete, quantized energy levels within the hydrogen atom.
549 Hard
549 of 592
Application +4 | -1
What is the de Broglie wavelength? Assume that one mole of protons has a mass equal to 1 g?1 \mathrm{~g}? (h=6.626×(h=6.626 \times 10−27erg s)10^{-27} \text{erg s})
View answer and explanation
Correct option: B
1.33×10−3A∘1.33 \times 10^{-3}\mathop {\text{A}}\limits^ \circ
Option B (1.33×10−3A∘1.33 \times 10^{-3}\mathop {\text{A}}\limits^ {\circ}) is correct because the de Broglie wavelength is calculated using the formula λ=hmv\lambda = \frac{{h}}{{mv}}, where m is the mass of one proton and v is the velocity.
550 Easy
550 of 592
Remembering +4 | -1
In case of dx2−y2d_{x^2-y^2} orbital:
View answer and explanation
Correct option: C
Probability of finding the electron is maximum along
x and y-axes
Option C (Probability of finding the electron is maximum along x and y-axes) is correct because the dx2−y2d_{x^2-y^2} orbital has lobes along the x and y axes, indicating a higher probability of finding electrons along those axes.
551 Easy
551 of 592
Remembering +4 | -1
The electronic energy levels of the hydrogen atom in the Bohr^' s theory are called:
View answer and explanation
Correct option: B
Orbits
Option B (Orbits) is correct because Bohr's theory postulates that electrons revolve around the nucleus in specific, quantized orbits.
552 Medium
552 of 592
Application +4 | -1
The principal quantum number of H-atom orbital, if the electron energy is -3.4 eV, will be
View answer and explanation
Correct option: B
2
Option B (2) is correct because the energy of an electron in a hydrogen atom is given by -13.6/n² eV, and solving for n when the energy is -3.4 eV gives n=2.
553 Easy
553 of 592
Application +4 | -1
If the electron in hydrogen atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by
View answer and explanation
Correct option: A
λ=365R\lambda = \frac{{36}}{{5R}}
Option A (λ=365R\lambda = \frac{{36}}{{5R}}) is correct because it is derived from the Rydberg formula for the wavelength of emitted radiation when an electron transitions between energy levels.
554 Medium
554 of 592
Application +4 | -1
A typical golf ball weighs 40.0 g40.0 \mathrm{~g}. If it is moving with a velocity of 20.0 ms−120.0 \mathrm{~ms}^{-1}, its de-Broglie wavelength is
View answer and explanation
Correct option: C
8.28×10−25 nm8.28 \times 10^{-25} \mathrm{~nm}
Option B (8.28×10−32 nm8.28 \times 10^{-32} \mathrm{~nm}) is correct because it is calculated using the de Broglie wavelength formula, λ=hmv\lambda = \frac{h}{mv}.
555 Easy
555 of 592
Understanding +4 | -1
Which is not an atomic orbital?
View answer and explanation
Correct option: A
2d
Option A (2d) is correct because the 2d orbital does not exist according to quantum mechanical principles.
556 Easy
556 of 592
Application +4 | -1
The energy in joule corresponding to light of wavelength 45 nm is
View answer and explanation
Correct option: D
4.42×10−184.42 \times {10^{ - 18}}
Option D (4.42×10−184.42 \times {10^{ - 18}}) is correct because the energy is calculated using the Planck's constant, speed of light, and the wavelength of the light.
557 Medium
557 of 592
Application +4 | -1
Bond dissociation on energy of Br2\mathrm{Br}_2 is 200 kJ/mole200 \mathrm{~kJ} / \mathrm{mole}. The longest wavelength of photon that can break this bond would be (NA×hc=0.12 J)\left(N_{\mathrm{A}} \times h c=0.12 \mathrm{~J}\right)
View answer and explanation
Correct option: C
6.0×10−7 m6.0 \times 10^{-7} \mathrm{~m}
Option C (6.0×10−7 m6.0 \times 10^{-7} \mathrm{~m}) is correct because it is the longest wavelength of a photon that can break the Br2Br_2 bond, calculated using the energy-wavelength relationship.
558 Medium
558 of 592
Application +4 | -1
According to Heisenberg's uncertainty principle, the product of uncertainties in position and velocities for an electron of mass 9.1×10−31 kg9.1 \times 10^{-31} {~kg} is
View answer and explanation
Correct option: C
5.8×10−5 m2 s−15.8 \times 10^{-5} {~m}^2 {~s}^{-1}
Option C (5.8×10−5 m2 s−15.8 \times 10^{-5} \mathrm{~m}^2 \mathrm{~s}^{-1}) is correct because it is calculated using Heisenberg's uncertainty principle.
559 Easy
559 of 592
Application +4 | -1
A metallic ion M2+{M^{2 + }} ion has a electronic configuration of 2, 8, 14 and the ionic weight is 56 amu. The number of neutrons in its nucleus are
View answer and explanation
Correct option: A
30
Option A (30) is correct because the number of neutrons is the difference between the mass number and the atomic number.
560 Hard
560 of 592
Application +4 | -1
Energy required to stop the ejection of electrons from Cu{Cu} plate is 0.24 eV0.24\, \mathrm{eV}. If radiation of λ=253.7 nm\lambda=253.7 \mathrm{~nm} strikes the plate, the work function is
View answer and explanation
Correct option: A
4.65 eV4.65\, \mathrm{eV}
Option A (4.65 eV4.65\, \mathrm{eV}) is correct because the work function is calculated by subtracting the stopping potential energy from the photon energy.
561 Medium
561 of 592
Application +4 | -1
Quantum numbers for some electrons are given below

A : n=4,I=1n=4, I=1\qquad B : n=4,I=0n=4, I=0\qquad C: n=3,I=2n=3, I=2\qquad D: n=3,I=1n=3, I=1

The correct increasing order of energy of electrons
View answer and explanation
Correct option: C
D << B << C << A
Option C (D << B << C << A) is correct because it represents the increasing order of energy based on the (n+l) rule and the n value when (n+l) is the same.
562 Medium
562 of 592
Application +4 | -1
The radius of an orbit of H – atom is 4.77×10−10m4.77 \times {10^{ - 10}}m. Then the angular momentum of an electron in this orbit is
View answer and explanation
Correct option: B
1.5h/π1.5h/\pi
Option B (1.5h/π1.5h/\pi) is correct because the radius is used to determine the orbit number (n=3), and then the angular momentum is calculated using the formula L=nh/2πL = nh/2\pi.
563 Easy
563 of 592
Remembering +4 | -1
Which among the following radiations has maximum frequency?
View answer and explanation
Correct option: C
γ\gamma-rays
Option C (γ\gamma-rays) is correct because gamma rays have the highest frequency among the given options in the electromagnetic spectrum.
564 Easy
564 of 592
Remembering +4 | -1
How many neutrons are present in tritium nucleus?
View answer and explanation
Correct option: A
2
Option A (2) is correct because tritium, an isotope of hydrogen, has 1 proton and 2 neutrons in its nucleus (mass number 3, atomic number 1).
565 Medium
565 of 592
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 quadrupling the energy reduces the wavelength to one-fourth.
566 Medium
566 of 592
Understanding +4 | -1
ASSERTION (A) : Two electrons in an atom can have the same values of four quantum numbers.

REASON (R) : Two electrons in an atom can be present in the same shell, subshell and orbital and have the same spin.
View answer and explanation
Correct option: D
If AA and RR both are wrong.
Option D (If A and R both are wrong) is correct because Assertion (A) is wrong due to Pauli Exclusion Principle, and Reason (R) is also wrong because electrons in the same orbital must have opposite spins.
567 Medium
567 of 592
Remembering +4 | -1
The principal and azimuthal quantum number of electrons in 4f4 f orbitals are
View answer and explanation
Correct option: C
4,3
Option C (4,3) is correct because for a 4f orbital, the principal quantum number (n) is 4 and the azimuthal quantum number (l) is 3.
568 Medium
568 of 592
Application +4 | -1
Based on the equation ΔE=−2.0×10−18 J(1n22−1n12)\Delta E=-2.0 \times 10^{-18} \mathrm{~J}\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)
the wavelength of the light that must be absorbed to excite hydrogen electron from level n=1n=1 to level n=2n=2 will be (h=6.625×10−34 J s,c=3×108 m s−1)\left(h=6.625 \times 10^{-34} \mathrm{~J} \mathrm{~s}, c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\right)
View answer and explanation
Correct option: A
1.325×10−7 m1.325 \times 10^{-7} \mathrm{~m}
Option A (1.325×10−7 m1.325 \times 10^{-7} \mathrm{~m}) is correct because it is the calculated wavelength of light absorbed during the electron transition from n=1 to n=2 in a hydrogen atom, using the provided energy equation.
569 Easy
569 of 592
Understanding +4 | -1
Which one of the following set of ions represents a collection of isoelectronic species
View answer and explanation
Correct option: B
CN−, C2−2, N2C{N^ - },\,C_2^{ - 2},\,{N_2}
Option B (CN−,C2−2,N2CN^-, C_2^{-2}, N_2) is correct because all three species have the same number of electrons (14).
570 Medium
570 of 592
Analyzing +4 | -1
Statement-I: Emitted radiation will be 0.408×10−190.408 \times 10^{-19} joule when an electron jump from n=4n=4 to n=2n=2 in H\mathrm{H}-atom.

Statement-II: Balmer series radiations belong to IR region for hydrogen atom only.
View answer and explanation
Correct option: B
Statement I and Statement II both are incorrect
The correct answer is B (Statement I and Statement II both are incorrect) because the emitted radiation energy is 4.08 × 10-19 J, and Balmer series radiations belong to the visible region.
Revision notes and key takeaways

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

Showing 541-570 of 592 questions
0 /592
0✓ 0✗
0% Accuracy
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.