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A charged particle $q$ of mass $m$ is projected along $y$ -axis at t=0 from the origin with a velocity $v_{0}$ . If an electric field $E$ also exists along the $x$ -axis, the time at which the de-Broglie wavelength of the particles becomes half of the initial value is

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$3 (\frac{m v_{0}}{q E})$

$\sqrt{3} (\frac{m v_{0}}{q E})$

$2 (\frac{m v_{0}}{q E})$

$\frac{m v_{0}}{q E}$

answer is C.

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Detailed Solution

Initial de-Broglie wavelength is $λ_{1} = \frac{h}{m v_{0}}$

At time $t$ , we have

$\vec{v} = v_{0} \hat{j} + (\frac{q E}{m}) t \hat{i}$

$\Rightarrow | \vec{v} | = \sqrt{v_{0}^{2} + \frac{q^{2} E^{2} t^{2}}{m^{2}}}$

$\Rightarrow λ_{2} = \frac{h}{m | \vec{v} |} = \frac{h}{\sqrt{m^{2} v_{0}^{2} + q^{2} E^{2} t^{2}}}$

Since $λ_{2} = \frac{λ_{1}}{2}$

$\Rightarrow \sqrt{m^{2} v_{0}^{2} + q^{2} E^{2} t^{2}} = 2 m v_{0}$

$\Rightarrow 3 m^{2} v_{0}^{2} = q^{2} E^{2} t^{2}$

$\Rightarrow t = \frac{\sqrt{3} m v_{0}}{q E}$

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