Question

In the nuclear transmutation, ${}_4^{9}Be+X\to {}_4^{8}Be+Y$ X and Y are

• A. $(\gamma ,n)$
• B. (p, D)
• C. (n, D)
• D. $(\gamma ,p)$

Answer 1

Answer: (A), (B)

(p, D)

${}_4^{9}Be+{}_a^{b}X\to {}_4^{8}Be+{}_c^{d}Y$
As we can see from the above reaction, the atomic number of the element remains the same, but the mass number differs.

We start by applying the law conservation of mass and conservation of energy (separately).

$\Rightarrow$ 4 + a = 4 + c
$\Rightarrow$ 9 + b = 8 + d
$\begin{array}{cc}IfX={}_0^0\gamma & a=0\\ & b=0\end{array}\begin{array}{cc}y={}_0^{1}n& c=0\\ & d=1\end{array}\begin{array}{cc}IfX={}_1^{1}P& a=1\\ & b=1\end{array}\begin{array}{cc}Y={}_D^2& c=1\\ & d=2\end{array}$

${}_4^{9}Be+{}_0^0\gamma \to {}_4^{8}Be+{}_0^{1}n{}^{}{}_4^{9}Be+{}_1^{1}p\to {}_4^{8}Be+{}_1^{2}D$