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. $\frac{h}{2m_{A^{V}}}\left\lbrack \frac{\left( m_{A} + m_{B} \right)}{\left( m_{A} - m_{B} \right)} - 1 \right\rbrack$
• B. )$\frac{h}{2m_{A^{V}}}\left\lbrack \frac{\left( m_{A} - m_{B} \right)}{\left( m_{A} + m_{B} \right)} - 1 \right\rbrack$
• C. )$\frac{h}{m_{A^{V}}}\left\lbrack \frac{\left( m_{A} + m_{B} \right)}{\left( m_{A} - m_{B} \right)} - 1 \right\rbrack$
• D. )$\frac{2h}{m_{A^{V}}}\left\lbrack \frac{\left( m_{A} + m_{B} \right)}{\left( m_{A} - m_{B} \right)} - 1 \right\rbrack$

Answer 1

Answer: (C)

)$\frac{h}{m_{A^{V}}}\left\lbrack \frac{\left( m_{A} + m_{B} \right)}{\left( m_{A} - m_{B} \right)} - 1 \right\rbrack$

Answer 2

: According to law of conservation of momentum

$m_{A}v + m_{B} \times 0 = m_{A}v_{A} + m_{B}v_{B}$

Or $m_{A}(v - v_{A}) = m_{B}v_{B}$……. (i)

According to law of conservation of kinetic 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

$v + v_{A} = v_{B}$…… (iii)

Solving (i) and (iii) we get

$v_{A} = \left( \frac{m_{A} - m_{B}}{m_{A} + m_{B}} \right)vandv_{B} = \left( \frac{2m_{A}}{m_{A} + m_{B}} \right)v$

$\lambda_{initial} = \frac{h}{m_{Av}}$

$\lambda_{final} = \frac{h}{m_{A}V_{A}} = \frac{h(m_{A} + m_{B})}{m_{A}(m_{A} - m_{B})v}$

$\therefore\Delta\lambda = \lambda_{final} - \lambda_{initial} = \frac{h}{m_{A}v}\left\lbrack \frac{(m_{A} + m_{B})}{m_{A} - m_{B}} - 1 \right\rbrack$