Unit 18: Atoms and Nuclei — Online MCQ Test
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Practice Unit 18: Atoms and Nuclei with a free chapter-wise online MCQ test.
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....
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Unit 18: Atoms and Nuclei — Important Questions & Answers
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).
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.
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.
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.
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.
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.
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.