A nucleus of mass M initially at rest splits into two fragments of masses m/3 and 2m/3. Find the ratio of de-Broglie wavelength of the two fragments.

Solution

Given, a molecule of mass M.
It splits into two fragments of masses m/3 and 2m/3.

Let,
$m_{1} = \frac{m}{3} and m_{2} = \frac{2 m}{3}$
But,
$m_{1} + m_{2} = M \Rightarrow m_{1} = M - m_{2}$

Following the law of conservation of linear momentum,
$m_{1} v_{1} + m_{2} v_{2} = 0$

$\Rightarrow$ $m_{1} v_{1} = - m_{2} v_{2}$

Also,De-broglie wavelength is, $λ = \frac{h}{mv}$

Hence, $|\frac{λ_{1}}{λ_{2}}| = |\frac{m_{2} v_{2}}{m_{1} v_{1}}| = 1 .$

Thus, the ratio of wavelength is 1.

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Wave Optics

A nucleus of mass M initially at rest splits into two fragments of masses m/3 and 2m/3. Find the ratio of de-Broglie wavelength of the two fragments.

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