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

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Solution

$\begin{array}{l} G i v e n : m_{p} = 200 \times 10^{6} e V / c^{2} a n d m_{H} = 10^{9} e V / c^{2} \\ D i v i d i n g, \\ \frac{m_{H}}{m_{p}} = \frac{10^{9}}{200 \times 10^{6}} = 5 \\ \Rightarrow m_{H} = 5 m_{p} \\ M a s s o f h y d r o g e n a t o m i s 5 t i m e s m a s s o f c o l l i d i n g p a r t i c l e . \end{array}$

$\begin{array}{l} C o n s e r v a t i o n o f M o m e n t u m, \\ m_{p} v_{p} = m_{H} v_{H} \\ \Rightarrow m v = 5 m v_{H} \\ \Rightarrow v_{H} = \frac{v}{5} \\ L o s s i n e n e r g y = K_{i} - K_{f} = \frac{1}{2} m v^{2} - \frac{1}{2} (5 m) {(\frac{v}{5})}^{2} = \frac{4}{10} m v^{2} \\ T h i s e n e r g y g o e s i n t o e x c i t i n g e l e c t r o n f r o m n = 1 t o n = 2 m e a n s △ E = 10.2 e V \\ \Rightarrow \frac{4}{10} m v^{2} = 10.2 e V \\ \Rightarrow \frac{1}{2} m v^{2} = 12.75 e V = \frac{51}{4} e V \end{array}$

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