Ncert Exercises & Solutions

A particle A with a mass m $_{A}$ is moving with a velocity v and hits particle B (mass m $_{B}$ ) at rest (one-dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat collision as elastic.

Hint: The de-Broglie wavelength depends on the momentum of the particle.

Step 1: Conserve the momentum.

As collision is elastic, hence laws of conservation of momentum and kinetic energy are obeyed.

According to the law of conservation of momentum,

$m_{A} v + m_{B} \times 0 = m_{A} v_{1} + m_{B} v_{2}$
$\Rightarrow m_{A} (v - v_{1}) = m_{B} v_{2}$

Step 2: Conserve the kinetic energy.

According to the law of conservation of kinetic energy,

$\begin{aligned} \frac{1}{2} m_{A} v^{2} & = \frac{1}{2} m_{A} v_{1}^{2} + \frac{1}{2} m_{B} v_{2}^{2} . . . (i) \\ \Rightarrow m_{A} (v - v_{1}^{2}) & = m_{8} v_{2}^{2} \\ \Rightarrow m_{A} (v - v_{1}) (v + v_{1}) & = m_{B} v_{2}^{2} . . . (i i) \end{aligned}$

Step 3: Find the final velocities of A and B.

Dividing Eq. (ii) by Eq. (i) ;

we get, $v + v_{1} = v_{2} o r v = v_{2} - v_{1} . . . (i i i)$

Step 4: Find the change in wavelength of particle A.

Solving Eqs. (i) and (ii), we get;

$v_{1} = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) v a n d v_{2} = (\frac{2 m_{A}}{m_{A} + m_{B}})$
$λ_{i n i t i a l} = \frac{h}{m_{A} v}$
$λ_{f i n a l} = \frac{h}{m_{A} v_{1}} = \frac{h (m_{A} + m_{B})}{m_{A} (m_{A} - m_{B}) v}$
$∆ λ = λ_{f i n a l} - λ_{i n i t i a l} = \frac{h}{m_{A} v} [\frac{m_{A} + m_{B}}{m_{A} - m_{B}} - 1]$

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