Chapter 7, Problem 77GP

To determine

Speed of the nucleus formed, after decay.

Answer to Problem 77GP

Velocity of the recoiled nucleus is $6.6\times 10^{3}m/s$

Explanation of Solution

Given:

Velocity of alpha particle is $3.8\times {10}^5\text{ m/s}$

Mass of recoiled nucleus is 57 times mass of parent nucleus.

Formula used:

Conservation of momentum

Initial Momentum=final momentum

  $m_p{v}_p=m_{\alpha }{v}_{\alpha }+m_R{v}_R$

Where.

P-parent,

  $\alpha$ represents alpha particle

R represents recoiled nucleus.

Calculation:

Consider general reaction of Alpha decay,

  $P\to R+\alpha$

  $m_p{v}_p=m_{\alpha }{v}_{\alpha }+m_R{v}_R$

As initially mass is at rest. So, initial momentum would be zero.

  $\begin{array}{l}0=m_{\alpha }\left(3.8\times 10^5\right)+57m_{\alpha }{v}_R\\ 0=3.8\times 10^5+57v_R\\ 3.8\times 10^5=57v_R\\ v_R=-\frac{3.8\times 10^5}{57}\\ v_R=-6.67\times 10^{3}m/s\end{array}$

Negative sign is neglected.

Conclusion:

Velocity of the recoiled nucleus is $6.67\times 10^{3}m/s$