Question

A nucleus with mass number $220$ initially at rest emits an alpha particle. If the Q value of the reaction is $5.5MeV$, the kinetic energy of the alpha particle is

• A. $5.6MeV$
• B. $4.4MeV$
• C. $6.5MeV$
• D. $5.4MeV$

Answer 1

The correct option is D

$5.4MeV$

Step 1: Given data

Q-factor, $Q=5.5MeV$

The mass number of the nucleus$=220$

Step 2: Formula used

By conservation of linear momentum,

$M_{\alpha }{v}_{\alpha }=M_y{v}_y$

$M_{\alpha },v_{\alpha }$ are mass and velocity of alpha particle

$M_y,v_y$ are mass and velocity of the daughter nuclei $Y$

Mathematically it is given by,$\begin{array}{l}{m}_1{u}_1+m_2{u}_2\end{array}$=$\begin{array}{l}{m}_1{v}_1+m_2{v}_2\end{array}$

Note:

Q-factor,

$Q={\left(KE\right)}_{\alpha }+{\left(KE\right)}_y$

The kinetic energy of the alpha particle is${\left(KE\right)}_{\alpha }$

${\left(KE\right)}_y$ is the kinetic energy of daughter nuclei $Y$

$Q$is the Q-factor

Step 3: Calculation
Let $X$ be the nucleus with the mass number $220$ and $Y$ be the daughter nuclei produced.

Hence, we have

${}^{220}X\to {}^{216}Y+{}_2{}^4\alpha$

$$\begin{gathered} M_{\alpha }=4 \\ {M}_y=216 \end{gathered}$$

Due to the conservation of linear momentum,

$$\begin{gathered} M_{\alpha }{v}_{\alpha }=M_y{v}_y \\ \Rightarrow v_y=\frac{M_{\alpha }{v}_{\alpha }}{M_y} \end{gathered}$$

Q-value = kinetic energy of alpha particle ($\alpha$) + kinetic energy of $Y$

$$\begin{gathered} Q={\left(KE\right)}_{\alpha }+{\left(KE\right)}_y \\ \Rightarrow Q=\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2+\frac{1}{2}{M}_y{v}_y^2 \\ \Rightarrow Q=\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2+\frac{1}{2}\frac{M_{\alpha }^2{v}_{\alpha }^2}{M_y} \\ \Rightarrow Q=\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2\left[1+\frac{M_{\alpha }}{M_y}\right] \\ \Rightarrow Q={\left(KE\right)}_{\alpha }\times \left[1+\frac{M_{\alpha }}{M_y}\right] \\ \Rightarrow 5.5={\left(KE\right)}_{\alpha }\times \left[1+\frac{4}{216}\right] \\ \Rightarrow 5.5={\left(KE\right)}_{\alpha }\times \left[\frac{220}{216}\right] \\ \Rightarrow {\left(KE\right)}_{\alpha }=\frac{5.5\times 216}{220} \\ \Rightarrow {\left(KE\right)}_{\alpha }=5.4MeV \end{gathered}$$

Therefore, the kinetic energy of the alpha particle is $5.4MeV$.

Option D is correct.