Chapter 7, Problem 11P

To determine

The speed of alpha particle after emission.

Answer to Problem 11P

The speed of alpha particle after emission is $4235\text{ m/s}$ .

Explanation of Solution

Given:

Mass of atomic nucleus, $m=222\text{ u}$

The mass of alpha particle, $m_{\alpha }=4\text{ u}$

Mass of atomic nucleus after emission is $m_2$ .

Speed of atomic nucleus before emission, $u=420\text{ m/s}$

Speed of atomic nucleus after emission, $v_2=350\text{ m/s}$

Formula used:

  $p_i=p_f$

Where, $p_i$ = momentum before collision

  $p_f$ = momentum after collision

Calculation:

The magnitude of mass of atomic nucleus after emission is calculated as,

  $\begin{array}{l}{m}_2=m_1-m_{\alpha }\\ m_2=222-4\\ m_2=218\text{ u}\end{array}$

According to principle of conservation of momentum,

  $\begin{array}{l}{p}_i=p_f\\ mv=m_1{v}_1+m_2{v}_2\\ v_1=\frac{mv-m_2{v}_2}{m_1}\\ v_1=\frac{\left(222\times 420\right)-\left(218\times 350\right)}{4u}\\ v_1=4235\text{ m/s}\end{array}$

Conclusion:

The speed of alpha particle after emission is $v_1=4235\text{ m/s}$ .