Question

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

Answer 1

Hint: Here, we have to find the initial kinetic energy of the particle. We have to first use the law of conservation of momentum to find the relation between the velocities of the hydrogen atom and the other particle. Then, we will use the relation of loss of kinetic energy which is equally gained by the hydrogen atom to jump to its next excited state. The energy required to jump to first excited state by hydrogen is $10.2\text{ }eV$.

Complete answer:
The question has already provided us with the mass of the hydrogen atom and the particle as follows:
${{m}_{H}}=\text{1 }\dfrac{GeV}{{{c}^{2}}}$ and ${{m}_{\text{particle}}}=200\text{ }\dfrac{MeV}{{{c}^{2}}}$

Now, by applying the law of conservation of momentum we find,
$\therefore m{{v}_{0}}+0=0+5mv$
$\Rightarrow v=\dfrac{{{v}_{0}}}{5}$--(i)
This is the relation between the two velocities of the hydrogen atom and the other particle.

We know that the loss of kinetic energy during the collision is the energy which is gained by the hydrogen atom to jump to its first excited state. Hence, let us first find the loss in K.E.
$\therefore \dfrac{1}{2}mv_{0}^{2}-\dfrac{1}{2}\times 5m\times {{v}^{2}}$

Substituting the value of $v$ from equation (i) we get,
$\Rightarrow \dfrac{1}{2}mv_{0}^{2}-\dfrac{1}{2}\times 5m\times {{\left( \dfrac{{{v}_{0}}}{5} \right)}^{2}}$

Simplifying the loss of K.E., and equating with the energy required by the hydrogen atom to its first excited state we get,
$\therefore \dfrac{4}{5}\left( \dfrac{mv_{0}^{2}}{2} \right)=10.2$
$\Rightarrow \dfrac{4}{5}k=10.2$
$\Rightarrow k=\dfrac{51}{4}$

Comparing this with variable given in the question as $\dfrac{N}{4}$ we get,
$N=51$

Therefore, the value of N is 51.

Note: It must be noted that according to the law of conservation of energy, energy can neither be created nor destroyed hence it transfers from one energy to other. So, the loss in K.E. during collision is transferred to the energy required by the hydrogen atom to jump to its first excited state.