Identification of Projectile and Ejectile in Beryllium Transmutation

To determine the correct pair(s) $(\text{X}, \text{Y})$ for the given nuclear reaction, we apply the laws of conservation of mass number ($A$) and atomic number ($Z$).

The nuclear reaction can be represented as: ${}_{4}^{9}\text{Be} + {}_{Z_X}^{A_X}\text{X} \longrightarrow {}_{4}^{8}\text{Be} + {}_{Z_Y}^{A_Y}\text{Y}$

1. Conservation of Mass Number ($A$): $9 + A_X = 8 + A_Y \implies A_Y - A_X = 1$

2. Conservation of Atomic Number ($Z$): $4 + Z_X = 4 + Z_Y \implies Z_X = Z_Y$

Now, we test each given option against these conservation laws:

• Option (A) $(\text{X}, \text{Y}) = (\gamma, n)$:

  • For gamma photon ($\gamma$): $A_\gamma = 0$, $Z_\gamma = 0$
  • For neutron ($n$): $A_n = 1$, $Z_n = 0$
  • Checking conservation laws:
    • $A_Y - A_X = 1 - 0 = 1$
    • $Z_X = Z_Y = 0$
  • Reaction equation: ${}_{4}^{9}\text{Be} + \gamma \longrightarrow {}_{4}^{8}\text{Be} + {}_{0}^{1}n$ Both $A$ and $Z$ are conserved. Hence, Option (A) is correct.

• Option (B) $(\text{X}, \text{Y}) = (p, D)$:

  • For proton ($p$ or ${}_{1}^{1}\text{H}$): $A_p = 1$, $Z_p = 1$
  • For deuteron ($D$ or ${}_{1}^{2}\text{H}$): $A_D = 2$, $Z_D = 1$
  • Checking conservation laws:
    • $A_Y - A_X = 2 - 1 = 1$
    • $Z_X = Z_Y = 1$
  • Reaction equation: ${}_{4}^{9}\text{Be} + {}_{1}^{1}p \longrightarrow {}_{4}^{8}\text{Be} + {}_{1}^{2}D$ Both $A$ and $Z$ are conserved. Hence, Option (B) is correct.

• Option (C) $(\text{X}, \text{Y}) = (n, D)$:

  • For neutron ($n$): $A_n = 1$, $Z_n = 0$
  • For deuteron ($D$): $A_D = 2$, $Z_D = 1$
  • Here, $Z_X = 0 \neq Z_Y = 1$, so atomic number is not conserved. Hence, Option (C) is incorrect.

• Option (D) $(\text{X}, \text{Y}) = (\gamma, p)$:

  • For gamma photon ($\gamma$): $A_\gamma = 0$, $Z_\gamma = 0$
  • For proton ($p$): $A_p = 1$, $Z_p = 1$
  • Here, $Z_X = 0 \neq Z_Y = 1$, so atomic number is not conserved. Hence, Option (D) is incorrect.

Therefore, the correct options are A and B.