Question

A particle of mass $200{\text{MeV/c}}^2$ collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in $\text{eV}$) is $\frac{N}{4}$. The value of $N$ is:

Answer 1

From linear momentum conservation,
$L_i=L_f$
$mv+0=0+5m{V}^{\prime }\Rightarrow V^{\prime }=\frac{v}{5}$

Loss of $\Delta KE=K{E}_i-K{E}_f=\frac{1}{2}m{v}^2-\frac{1}{2}\left(5m\right){\left(\frac{v}{5}\right)}^2$
$=\frac{1}{2}m{v}^2(1-\frac{1}{5})=\frac{4}{5}\left(\frac{m{v}^2}{2}\right)$
$=\frac{4}{5}K{E}_i=10.2\text{eV}$
[$\because$ Energy in first excited state of atom $=10.2\text{eV}$]
$K{E}_i=12.75\text{eV}=\frac{N}{4}\Rightarrow N=51$
The value of $N=51$