JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Chemical Kinetics

For the nuclear transmutation ${}_4^9\mathrm{Be}+\mathbf{X}⟶{}_4^8\mathrm{Be}+\mathbf{Y}$, we check mass and atomic number conservation:

(A) ${}_4^9\mathrm{Be}+\gamma \to {}_4^8\mathrm{Be}+{}_0^{1}n$: Mass 9 = 8 + 1 ✓, atomic number 4 = 4 + 0 ✓. Possible.

(B) ${}_4^9\mathrm{Be}+{}_1^{1}p\to {}_4^8\mathrm{Be}+{}_1^{2}D$: Mass 9 + 1 = 8 + 2 ✓, atomic number 4 + 1 = 4 + 1 ✓. Possible.

(C) ${}_4^9\mathrm{Be}+{}_0^{1}n\to {}_4^8\mathrm{Be}+{}_1^{2}D$: Mass 9 + 1 = 10 ≠ 8 + 2 = 10, but atomic number 4 + 0 = 4 ≠ 4 + 1 = 5. Not possible.

(D) ${}_4^9\mathrm{Be}+\gamma \to {}_4^8\mathrm{Be}+{}_1^{1}p$: Atomic number 4 ≠ 4 + 1. Not possible.