Match List-I with List-II
| List-I (Nuclear Decay) | List-II (Emission of particles) | ||
|---|---|---|---|
| (A) | \( {}^{238}_{92}\mathrm{U}\rightarrow{}^{234}_{91}\mathrm{Pa} \) | (I) | One \( \alpha \) particle and one \( \beta^{+} \) particle |
| (B) | \( {}^{214}_{82}\mathrm{Pb}\rightarrow{}^{210}_{82}\mathrm{Pb} \) | (II) | Three \( \beta^{-} \) particles and one \( \alpha \) particle |
| (C) | \( {}^{210}_{81}\mathrm{Tl}\rightarrow{}^{206}_{82}\mathrm{Pb} \) | (III) | Two \( \beta^{-} \) particles and one \( \alpha \) particle |
| (D) | \( {}^{228}_{91}\mathrm{Pa}\rightarrow{}^{224}_{88}\mathrm{Ra} \) | (IV) | One \( \alpha \) particle and one \( \beta^{-} \) particle |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
In alpha decay A falls by 4 and Z by 2; in beta-minus decay Z rises by 1 and in beta-plus decay Z falls by 1, with A unchanged in both. In every decay here A falls by 4, so each involves one alpha particle; the change in Z then fixes the beta particles.
(A) \( {}^{238}_{92}\mathrm{U}\rightarrow{}^{234}_{91}\mathrm{Pa} \) — this matches (IV) One \( \alpha \) particle and one \( \beta^{-} \) particle. After the alpha decay Z = 90; reaching 91 needs one \( \beta^{-} \).
\( Z: \ 92 - 2 + 1 = 91 \)
(B) \( {}^{214}_{82}\mathrm{Pb}\rightarrow{}^{210}_{82}\mathrm{Pb} \) — this matches (III) Two \( \beta^{-} \) particles and one \( \alpha \) particle. The alpha decay gives Z = 80; returning to 82 needs two \( \beta^{-} \).
\( Z: \ 82 - 2 + 2 = 82 \)
(C) \( {}^{210}_{81}\mathrm{Tl}\rightarrow{}^{206}_{82}\mathrm{Pb} \) — this matches (II) Three \( \beta^{-} \) particles and one \( \alpha \) particle. The alpha decay gives Z = 79; reaching 82 needs three \( \beta^{-} \).
\( Z: \ 81 - 2 + 3 = 82 \)
(D) \( {}^{228}_{91}\mathrm{Pa}\rightarrow{}^{224}_{88}\mathrm{Ra} \) — this matches (I) One \( \alpha \) particle and one \( \beta^{+} \) particle. The alpha decay gives Z = 89; falling to 88 needs one \( \beta^{+} \).
\( Z: \ 91 - 2 - 1 = 88 \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Alpha Decay – “… In this process, it is observed that since \( {}^{4}_{2}\mathrm{He} \) contains two protons and two neutrons, the mass number and the atomic number of the daughter nucleus decreases by four and two, respectively.”
Beta Decay – “… In beta-minus decay, a neutron transforms into a proton within the nucleus according to
\( \displaystyle n \rightarrow p + e^{-} + \bar{\nu} \qquad (13.24) \)
whereas in beta-plus decay, a proton transforms into neutron (inside the nucleus) via
\( \displaystyle p \rightarrow n + e^{+} + \nu \qquad (13.25) \)
These processes show why the mass number A of a nuclide undergoing beta decay does not change; one of its constituent nucleons simply changes its character according to Eq. (13.24) or (13.25).”