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A heavy nucleus at rest breaks into two fragments, which fly off with velocities in the ratio $8:1$. What is the ratio of de-Broglie wavelengths of the fragments?
A) $1:2$
B) $1:8$
C) $4:1$
D $\text{None of these}$

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Last updated date: 23rd Feb 2024
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Hint: Whenever a body is in motion, its energies and all other quantities change, but the momentum remains constant. Hence, from the ratio of the velocities of the two fragments, the ratio of the masses of the two fragments can be obtained by conserving momentum.

Complete step by step solution:
Light behaves like a particle and also a wave depending on the experiment conducted. The photoelectric effect shows particle-like behaviour and travels in small packets called photons, whereas in Young’s double-slit experiment, it behaves as a wave and causes constructive and destructive interference at different points on the screen.
The scientist Louis de-Broglie suggested that even electrons possess both particle-like and wave-like properties and should be associated with the equation as that of the photons, i.e., ${\lambda _e} = \dfrac{h}{{mv}}$ .
Davisson and Germer conducted an experiment where they shot electrons just like the photons to check for electrons’ wave nature. They found that the electrons did not just create two bright spots but actually exhibited constructive and destructive interference at different points on the screen, just like the photons. Hence, de Broglie's postulates were proved.
By conservation of momentum $(p = mv)$:
$\Delta p = 0$
$ \Rightarrow {m_1}{v_1} = {m_2}{v_2}$
Given that $\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{8}{1}$………….. equation (1)
$ \Rightarrow \dfrac{{{m_2}}}{{{m_1}}} = \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{8}{1}$
$\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{8}$...…….. equation (2)
The de-Broglie wavelength $(\lambda )$ is expressed as:
$\lambda = \dfrac{h}{{mv}}$
where,
$h = $ Planck’s constant $ = 6.626 \times {10^{ - 34}}{m^2}Kg.{s^{ - 1}}$
$m = $ mass of particle
$v = $ velocity of particle
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{h}{{{m_1}{v_1}}} \times \dfrac{{{m_2}{v_2}}}{h}$
Canceling $h$ from both sides:
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{m_2}}}{{{m_1}}} \times \dfrac{{{v_2}}}{{{v_1}}}$
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{8}{1} \times \dfrac{1}{8}$
$\therefore \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{1}{1}$

The correct answer is [D], None of these.

Note: Planck’s constant is the link between the macroscopic and the microscopic worlds. Matter particles as waves can be recognized with the help of the emergence of the Planck’s constant. In all, the postulates of de-Broglie influenced fundamental quantum physics.