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A particle A ...

Question

A particle A with a mass $m_{A}$ is moving with a velocity v and hits a particle B of mass $m_{B}$ at rest. If motion is one dimensional and take the collision is elastic, then the change in the de Broglie wavelength of the particle A is

A

\frac{h}{2 m_{A} v} [\frac{(m_{A} + m_{B})}{(m_{A} - m_{B})} - 1]

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B

\frac{h}{m_{A} v} [\frac{(m_{A} - m_{B})}{(m_{A} + m_{B})} - 1]

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C

\frac{h}{m_{A} v} [\frac{(m_{A} + m_{B})}{(m_{A} - m_{B})} - 1]

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D

\frac{2 h}{m_{A} v} [\frac{(m_{A} + m_{B})}{(m_{A} - m_{B})} + 1]

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Solution

The correct option is C $\frac{h}{m_{A} v} [\frac{(m_{A} + m_{B})}{(m_{A} - m_{B})} - 1]$
According to law of conversation of momentum
$m_{A} v + m_{B} \times 0 = m_{A} v_{A} + m_{B} v_{B}$
or $m_{A} (v - v_{B}) = m_{B} v_{B}$ ....(i)
According to law of conversation of klinetic energy
$\frac{1}{2} m_{A} v^{2} = \frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} v_{B}^{2}$

or $m_{A} (v^{2} - V_{A}^{2}) = m_{B} v_{B}^{2}$

or $m_{A} (v - v_{A}) (v + v_{A}) = m_{B} v_{b}^{2}$ ...(ii)

Dividing (ii) by (i), we get
$v + v_{A} = v_{B} o r v = v_{B} - v_{A}$ ...(iii)

Solving (i) and (iii), we get
$v_{A} = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) v$ and $v_{B} = (\frac{2 m_{A}}{m_{A} + m_{B}}) v$

$λ_{i n i t i a l} = \frac{h}{m_{A} v}$ and $λ_{f i n a l} = \frac{h}{m_{A} v_{A}} = \frac{h (m_{A} + m_{B})}{m_{A} (m_{A} - m_{B}) v}$

$∴ Δ λ = λ_{f i n a l} - λ_{i n i t i a l} = \frac{h}{m_{A} v} [\frac{(m_{A} + m_{B})}{(m_{A} - m_{B})} - 1]$

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