A particle of mass $200 \frac{M e V}{c^{2}}$ 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 $(i n e V)$ is $\frac{N}{4}$ . The value of $N$ is : (Given the mass of the hydrogen atom to be $1 \frac{G e V}{c^{2}}$ )

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Solution

Step1: Given data.
Mass of the particle, $m_{p} = 200 \frac{M e V}{c^{2}}$
Mass of the hydrogen atom, $m_{h} = 1 \frac{G e V}{c^{2}} = 1000 \frac{M e V}{c^{2}}$
Initial kinetic energy, $K E_{i} = \frac{N}{4}$
Step2: Finding the value of N in the initial kinetic energy of the particle.
We know that,
From law of conservation of linear momentum,
$L_{i} = L_{f}$
Where, $L_{i}$ linear momentum before collision and $L_{f}$ linear momentum after collision.
$m_{p} v_{p} + m_{h} v_{h} = m_{p} v'_{p} + m_{h} v'_{h}$
Where $v_{p}$ particle velocity and $v_{h}$ hydrogen velocity.
$m_{p} v_{p} + 0 = 0 + m_{h} v'_{h}$
$\Rightarrow$ $v'_{h} = \frac{m}{5 m} v_{p} = \frac{v_{p}}{5}$
Since
$v'_{h} = \frac{v}{5}$ …… $(i)$
Now,
Loss of kinetic energy,
$K E_{l o s s} = K E_{i} - K E_{f}$
$\Rightarrow$ $K E_{l o s s} = \frac{1}{2} m v_{p}^{2} - \frac{1}{2} 5 m v'_{h}^{2}$
$\Rightarrow$ $K E_{l o s s} = \frac{1}{2} m v^{2} - \frac{1}{2} 5 m {(\frac{v}{5})}^{2}$ $[∵ (i)]$
$\Rightarrow$ $K E_{l o s s} = \frac{1}{2} m v^{2} (1 - \frac{1}{5})$
$\Rightarrow$ $K E_{l o s s} = \frac{4}{5} (\frac{m v^{2}}{2})$
$\Rightarrow$ $K E_{l o s s} = \frac{4}{5} K E_{i}$
Also we know that the energy in first excited state of atom is $10.2 e V$
Therefore,
$\Rightarrow$ $10.2 e V = \frac{4}{5} K E_{i}$
$\Rightarrow$ $K E_{i} = 12.75 e V$
$\Rightarrow$ $\frac{N}{4} = 12.75 e V$ $[∵ K E_{i} = \frac{N}{4}]$
$\Rightarrow$ $N = 51$
Hence the value of $N$ is $51$

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