An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?

Solution

Given, de-broglie wavelength assosciated with electron and an alpha particle is same.

Now,

Kinetic energy, $K . E . = \frac{1}{2} {mv}^{2} = \frac{1}{2} m \frac{p^{2}}{m^{2}}$

$\Rightarrow$ $K . E . = \frac{p^{2}}{2 m}$ ... (1) $(∵ p = mv)$

and,
$λ_{e} = λ_{α} \frac{h}{p_{e}} = \frac{h}{p_{α}}$

where, p_e and $p_{α}$ is the momentum of electron and alpha particle respectively.

So, $p_{e}^{2} = p_{α}^{2} - I$

From equation (1)

$\frac{K . E ._{e}}{K . E ._{α}} = \frac{\frac{1}{2} \frac{{p_{e}}^{2}}{2 m_{e}}}{\frac{1}{2} \frac{{ρ_{α}}^{2}}{2 m_{α}}}$

$∴$ $\frac{K . E ._{e}}{K . E ._{α}} = \frac{m_{α}}{m_{e}} \times \frac{{p_{e}}^{2}}{p_{α}} = \frac{m_{α}}{m_{e}} .$

is the required relation.

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Wave Optics

An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?

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