A particle of mass

Question:

A particle of mass $200 \mathrm{MeV} / \mathrm{c}^{2}$ 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 $\mathrm{eV}$ ) is $\frac{\mathrm{N}}{4}$. The value of $\mathrm{N}$ is : (Given the mass of the hydrogen atom to be $1 \mathrm{GeV} / \mathrm{c}^{2}$ )

Solution:

$\mathrm{mV}_{0}=\mathrm{MV}=\mathrm{p}$

$10.2=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2 \mathrm{M}}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}\left(1-\frac{\mathrm{m}}{\mathrm{M}}\right)$

$=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}(1-0.2)$

$\Rightarrow \frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\mathrm{K}=\frac{10.2}{0.8}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now