Question

A nucleus X initially at rest, undergoes alpha decay according to the equation ${}^{232}{X}_{92}\to {}^A{Y}_{90}+\alpha$. What fraction of total energy released in the decay will be the kinetic energy of the alpha particle?

• A. $\frac{90}{92}$
• B. $\frac{228}{232}$
• C. $\sqrt{\frac{228}{232}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{228}{232}$

$\begin{array}{c}\text{Given-}{\ast }^{232}{\times }_{92}{\to }^A{Y}_{90}+\alpha \\ \text{we know that emission of one}\alpha \text{particle reduces}\\ \text{atomic number by}2\text{and atomic mass by}4.\end{array}$

$\Rightarrow {}^{232}{X}_{92}\to {}^{228}{Y}_{90}+\alpha \left[{}_2^{4}He\right)$

$\begin{array}{c}\text{Let the two nuclei gets separated-}\\ \Rightarrow \text{By momentum conservation-}\\ \begin{array}{cc}\text{we have}& 228m_0{V}_1=4m_0{V}_2\\ & \Rightarrow v_2=57v_1---1\end{array}\end{array}$

$\begin{array}{c}\text{Total}kE=\frac{1}{2}\cdot 228m_0\cdot v_1^2+\frac{1}{2}\cdot 4m_0{v}_2^2\\ \text{KE of}\alpha \text{particle}=\frac{1}{2}\cdot 4m_0\cdot v_2^2\\ \text{Fraction of}kE\text{of}\alpha \text{particle}=(\frac{1}{2}\cdot 4m_0\cdot v_2^2)/(\frac{1}{2}\cdot 228m0\cdot v_1^2+\frac{1}{2}\cdot 4m_0\cdot v_2^2)\end{array}$

$\begin{array}{c}\text{Foom equation}-1\text{,we get -}\\ \text{fraction of}kE\text{of}\alpha \text{purticle}=\frac{228}{232}\end{array}$