Q123 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
When ${ }^{10} \mathrm{~B}_5$ nuclei are bombarded by neutrons, one of the resultant nuclei is ${ }^7 \mathrm{Li}_3$. Then the emitted particle will be:
- ABeta particle
- BGamma particle
- CNeutrons
- Correct answerDAlpha particle
Solution
The nuclear reaction can be represented as follows:
${}^{10}_{5}\text{B} + {}^{1}_{0}\text{n} \rightarrow {}^{7}_{3}\text{Li} + {}^{A}_{Z}\text{X}$
Applying the law of conservation of mass number (A):
$10 + 1 = 7 + A \Rightarrow 11 = 7 + A \Rightarrow A = 4$
Applying the law of conservation of atomic number (Z):
$5 + 0 = 3 + Z \Rightarrow 5 = 3 + Z \Rightarrow Z = 2$
The particle with mass number 4 and atomic number 2 is the alpha particle (${}^{4}_{2}\text{He}$ or $\alpha$).
Answer: Alpha particle
${}^{10}_{5}\text{B} + {}^{1}_{0}\text{n} \rightarrow {}^{7}_{3}\text{Li} + {}^{A}_{Z}\text{X}$
Applying the law of conservation of mass number (A):
$10 + 1 = 7 + A \Rightarrow 11 = 7 + A \Rightarrow A = 4$
Applying the law of conservation of atomic number (Z):
$5 + 0 = 3 + Z \Rightarrow 5 = 3 + Z \Rightarrow Z = 2$
The particle with mass number 4 and atomic number 2 is the alpha particle (${}^{4}_{2}\text{He}$ or $\alpha$).
Answer: Alpha particle
Hence, the correct answer is option: 'D'.