An electron travels with a velocity of x $m s^{- 1}$ . For a proton to have the same de-broglie wavelength, the approximate velocity should be:
(Given, $\frac{m_{p}}{m_{e}} = 1840)$

\frac{1840}{x}

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\frac{x}{1840}

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1840 x

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Solution

The correct option is B $\frac{x}{1840}$
From de-Broglie's hypothesis,
$λ = \frac{h}{m_{p} v_{p}} = \frac{h}{m_{e} x}$
$λ$ = wavelength
m = mass of the particle
h = planck's constant
v = velocity of the particle.
At the same wavelength,
$m \propto \frac{1}{v} W e k n o w t h a t, \frac{m_{p}}{m_{e}} = 1840 m_{p} = 1840 m_{e} \frac{1}{v_{p}} = 1840 \frac{1}{v_{e}} ∴ V e l o c i t y o f p r o t o n = \frac{v_{e}}{1840} = \frac{x}{1840}$

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