Question

A U^(238) nucleus originally at rest decays by emitting an alpha -particle with a speed of vm/s .The recoil speed of residual nucleus in m/sec is​

Answer 1

Given :

•Mass of alpha particle is $\bf{m_{a}=4m}$

•The recoil speed of residual nucleus in m/sec=?

To Find :

•The recoil speed of residual nucleus in m/sec=?

Formula :

$\\ \blue\bigstar\boxed{\bf{m_{a}v_{a}+m_{r}v_{r}=0}}$

Solution :

⇒Mass of alpha particle is $\bf{m_{a}=4m}$

, where m is the mass of the photon.

And the mass of residual nucleus will be $\bf{m_{r}=(238-4)m =234 m}$

Momentum conservation$\bf{m_{a}v_{a}+m_{r}v_{r}=0}$

•Putting values of massses and $\bf{v_{a}=v}$

$\\ \implies\sf{4m \times v+ 234m \times v_{r}=0}$

$\\ \implies\boxed{\sf{v_{r}=\dfrac{-4v}{234}}}$

The recoil speed of residual nucleus in m/sec is

$\boxed{\sf{v_{r}=\dfrac{-4v}{234}}}$