First slide

Wave nature of matter

Question

A bomb projected from ground at an angle $θ (θ \neq 90 °)$ with horizontal explodes into two fragments of equal mass at top most point of its trajectory. If one of the fragment returns to point of projection then ratio of de-Broglie wavelength of second fragment just after explosion to bomb just before explosion is :

Moderate

Solution

From the law of conservation of momentum
$\begin{array}{l} m u \cos θ = - \frac{m}{2} u \cos θ + \frac{m}{2} v_{1} \\ v_{1} = 3 u \cos θ \dots (1) \\ λ_{D} = \frac{h}{m v} \dots (2) \end{array}$
The ratio of De-Broglie wavelength
$\Rightarrow \frac{(λ_{D})}{{(λ_{D})}_{2}} = \frac{\frac{h}{\frac{m}{2} (3 u \cos θ)}}{\frac{h}{m u \cos θ}} = \frac{2}{3}$
Therefore, the correct answer is 0.67.

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