Chapter Exam

Unit 18: Atoms and Nuclei — Online MCQ Test

Q: When a uranium isotope 235U is bombarded with a neutron, it generates 92Kr, three neutrons and:

Answer: A. 141Ba — In nuclear fission of 235U, the sum of atomic numbers and mass numbers must be conserved. 235 + 1 = 92 + X + 3(1). For 92Kr (Z=36), the remaining mass is 236 - 92 - 3 = 141. The balanced product is 141Ba (Z=56).

Q: A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h/e) where n is an integer, h is Planck's constant and e is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (m is the mass of the electron)

Answer: B. he/2m pi — Using flux quantization Phi = B*pi*r^2 = nh/e. For n=1, r^2 = h/(e*pi*B). M = I*A = (ev/2pir)*pi*r^2 = evr/2. Using Larmor frequency v/r = eB/m, M = e(eB/m * r^2)/2 = he/2m pi.

Q: De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

Answer: B. 0.67 nm — From Bohr's quantization condition, 2*pi*r = n*lambda. For n=2, r = n^2 * a0 = 4 * 0.052 nm = 0.208 nm. So, lambda = 2*pi*r / n = 2*3.14*0.208 / 2 = 3.14 * 0.208 ≈ 0.65 nm. Closest value is 0.67 nm.

Q: The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is

Answer: D. 1/4 — 1/λ = R(1/n1^2 - 1/n2^2). Transition 1: 1/λ1 = R(1/4 - 1/9) = 5R/36. λ1 = 36/5R. Transition 2: 1/λ2 = R(1/16 - 1/36) = (9-4)R/144 = 5R/144. λ2 = 144/5R. λ1/λ2 = 36/144 = 1/4.

Q: Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

Answer: B. 0.25 × 10⁷ m⁻¹ — For the last line of the Balmer series, n1 = 2 and n2 = ∞. The wave number 1/λ = R(1/n1² - 1/n2²) = 10⁷(1/4 - 0) = 0.25 × 10⁷ m⁻¹.

Q: Match the List-I with List-II. List-I: Spectral lines of hydrogen for transitions from (A) n = 3 to 2 (B) n = 4 to 2 (C) n = 5 to 2 (D) n = 6 to 2 List-II: Wavelengths (nm) I. 410.2 II. 434.1 III. 656.3 IV. 486.1 Choose the correct answer from the options given below:

Answer: D. A-III, B-IV, C-II, D-I — These are Balmer lines: n=3→2 is H-alpha 656.3 nm, n=4→2 is 486.1 nm, n=5→2 is 434.1 nm, and n=6→2 is 410.2 nm.

Q: Match the List-I with List-II. List-I (Spectral lines of hydrogen for transitions from) (A) n = 3 to 2 (B) n = 4 to 2 (C) n = 5 to 2 (D) n = 6 to 2 List-II (Wavelengths in nm) (I) 410.2 (II) 434.1 (III) 656.3 (IV) 486.1 Choose the correct answer from the options given below:

Answer: A. A-IV, B-III, C-I, D-II — These are Balmer series lines: 3→2 = 656.3 nm (Hα), 4→2 = 486.1 nm, 5→2 = 434.1 nm, 6→2 = 410.2 nm. Hence A-IV, B-III, C-II, D-I? The printed matching in the OCR is distorted; the correct correspondence is A-III, B-IV, C-II, D-I. Therefore option (4) is correct.

Q: In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:

Answer: D. 286, 80 — The chain starts from A = 290, Z = 82. Two α-decays reduce A by 8 and Z by 4, and two β^- decays increase Z by 2. Net result: A = 286 and Z = 80.

Q: For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is

Answer: C. 20 — Initially N₀ = 600. After disintegration of 450 nuclei, remaining nuclei = 600 – 450 = 150 = N₀/4. Since N = N₀(1/2)^(t/T½), we have 1/4 = (1/2)^(t/10), so t/10 = 2, giving t = 20 minutes.

Q: The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is

Answer: B. 1 : –1 — In a Bohr orbit, KE = –(Total Energy) and PE = 2×(Total Energy). So KE/TE = –TE/TE = –1, i.e., KE : TE = 1 : –1.

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Practice Unit 18: Atoms and Nuclei with a free chapter-wise online MCQ test for NEET NEET (All) PHYSICS. This chapter covers: Alpha-particle scattering experiment; Rutherford's model of atom; Bohr model energy levels hydrogen spectrum. Composition and size of nucleus atomic masses Mass-energy relation mas.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 18: Atoms and Nuclei — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) PHYSICS — Unit 18: Atoms and Nuclei, with answers and explanations. These are sample questions; the exam has its own separate question set.

When a uranium isotope 235U is bombarded with a neutron, it generates 92Kr, three neutrons and:

A. 141Ba ✓
B. 142Zr
C. 141Kr
D. 142Kr

Answer: A. 141Ba
In nuclear fission of 235U, the sum of atomic numbers and mass numbers must be conserved. 235 + 1 = 92 + X + 3(1). For 92Kr (Z=36), the remaining mass is 236 - 92 - 3 = 141. The balanced product is 141Ba (Z=56).

A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h/e) where n is an integer, h is Planck's constant and e is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (m is the mass of the electron)

A. he/m pi
B. he/2m pi ✓
C. heB/m pi
D. heB/2m pi

Answer: B. he/2m pi
Using flux quantization Phi = B*pi*r^2 = nh/e. For n=1, r^2 = h/(e*pi*B). M = I*A = (ev/2pir)*pi*r^2 = evr/2. Using Larmor frequency v/r = eB/m, M = e(eB/m * r^2)/2 = he/2m pi.

De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

A. 0.067 nm
B. 0.67 nm ✓
C. 1.67 nm
D. 2.67 nm

Answer: B. 0.67 nm
From Bohr's quantization condition, 2*pi*r = n*lambda. For n=2, r = n^2 * a0 = 4 * 0.052 nm = 0.208 nm. So, lambda = 2*pi*r / n = 2*3.14*0.208 / 2 = 3.14 * 0.208 ≈ 0.65 nm. Closest value is 0.67 nm.

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is

A. 1/36
B. 1/16
C. 1/9
D. 1/4 ✓

Answer: D. 1/4
1/λ = R(1/n1^2 - 1/n2^2). Transition 1: 1/λ1 = R(1/4 - 1/9) = 5R/36. λ1 = 36/5R. Transition 2: 1/λ2 = R(1/16 - 1/36) = (9-4)R/144 = 5R/144. λ2 = 144/5R. λ1/λ2 = 36/144 = 1/4.

Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

A. 0.5 × 10⁷ m⁻¹
B. 0.25 × 10⁷ m⁻¹ ✓
C. 2.5 × 10⁷ m⁻¹
D. 0.025 × 10⁴ m⁻¹

Answer: B. 0.25 × 10⁷ m⁻¹
For the last line of the Balmer series, n1 = 2 and n2 = ∞. The wave number 1/λ = R(1/n1² - 1/n2²) = 10⁷(1/4 - 0) = 0.25 × 10⁷ m⁻¹.

Match the List-I with List-II. List-I: Spectral lines of hydrogen for transitions from (A) n = 3 to 2 (B) n = 4 to 2 (C) n = 5 to 2 (D) n = 6 to 2 List-II: Wavelengths (nm) I. 410.2 II. 434.1 III. 656.3 IV. 486.1 Choose the correct answer from the options given below:

A. A-IV, B-III, C-I, D-II
B. A-I, B-II, C-III, D-IV
C. A-II, B-I, C-IV, D-III
D. A-III, B-IV, C-II, D-I ✓

Answer: D. A-III, B-IV, C-II, D-I
These are Balmer lines: n=3→2 is H-alpha 656.3 nm, n=4→2 is 486.1 nm, n=5→2 is 434.1 nm, and n=6→2 is 410.2 nm.

Match the List-I with List-II. List-I (Spectral lines of hydrogen for transitions from) (A) n = 3 to 2 (B) n = 4 to 2 (C) n = 5 to 2 (D) n = 6 to 2 List-II (Wavelengths in nm) (I) 410.2 (II) 434.1 (III) 656.3 (IV) 486.1 Choose the correct answer from the options given below:

A. A-IV, B-III, C-I, D-II ✓
B. A-I, B-II, C-III, D-IV
C. A-II, B-I, C-IV, D-III
D. A-III, B-IV, C-II, D-I

Answer: A. A-IV, B-III, C-I, D-II
These are Balmer series lines: 3→2 = 656.3 nm (Hα), 4→2 = 486.1 nm, 5→2 = 434.1 nm, 6→2 = 410.2 nm. Hence A-IV, B-III, C-II, D-I? The printed matching in the OCR is distorted; the correct correspondence is A-III, B-IV, C-II, D-I. Therefore option (4) is correct.

In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:

A. 288, 82
B. 286, 81
C. 280, 81
D. 286, 80 ✓

Answer: D. 286, 80
The chain starts from A = 290, Z = 82. Two α-decays reduce A by 8 and Z by 4, and two β^- decays increase Z by 2. Net result: A = 286 and Z = 80.

For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is

A. 30
B. 10
C. 20 ✓
D. 15

Answer: C. 20
Initially N₀ = 600. After disintegration of 450 nuclei, remaining nuclei = 600 – 450 = 150 = N₀/4. Since N = N₀(1/2)^(t/T½), we have 1/4 = (1/2)^(t/10), so t/10 = 2, giving t = 20 minutes.

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is