Q160 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
A radioactive nucleus ${}_{92}X^{235}$ decays to ${}_{91}Y^{231}$. Which of the following particles are emitted?
- Correct answerAOne alpha and one beta particle
- BTwo deuterons and one positron
- COne alpha and one proton
- DOne proton and four neutrons
Solution
The decay reaction can be written as:
${}_{92}X^{235} \rightarrow {}_{91}Y^{231} + n_{\alpha} ({}_{2}He^{4}) + n_{\beta} ({}_{-1}e^{0})$
By conservation of mass number:
$235 = 231 + 4n_{\alpha} + 0$
$4n_{\alpha} = 4 \Rightarrow n_{\alpha} = 1$
By conservation of atomic number:
$92 = 91 + 2n_{\alpha} - n_{\beta}$
$92 = 91 + 2(1) - n_{\beta}$
$92 = 93 - n_{\beta} \Rightarrow n_{\beta} = 1$
Thus, one alpha particle and one beta particle are emitted.
Answer: One alpha and one beta particle
${}_{92}X^{235} \rightarrow {}_{91}Y^{231} + n_{\alpha} ({}_{2}He^{4}) + n_{\beta} ({}_{-1}e^{0})$
By conservation of mass number:
$235 = 231 + 4n_{\alpha} + 0$
$4n_{\alpha} = 4 \Rightarrow n_{\alpha} = 1$
By conservation of atomic number:
$92 = 91 + 2n_{\alpha} - n_{\beta}$
$92 = 91 + 2(1) - n_{\beta}$
$92 = 93 - n_{\beta} \Rightarrow n_{\beta} = 1$
Thus, one alpha particle and one beta particle are emitted.
Answer: One alpha and one beta particle
Hence, the correct answer is option: 'A'.