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A particle of charge $\mathrm{q}$ and mass $\mathrm{m}$ enters a region of a transverse electric field of $\mathrm{E}_{0} \hat{\mathrm{j}}$ with initial velocity $v_{0} \hat{i}$. The time taken for the change in the de Broglie wavelength of the charge from the initial value of $\lambda_{0}$ to $\lambda_{0} / 3$ is proportional to
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The correct answer is:
$\frac{\mathrm{m}}{\mathrm{q}}$

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \Rightarrow \lambda \propto \frac{1}{\mathrm{v}}$
$\mathrm{v}=\sqrt{\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{0}^{2}}$
$\mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{y} \mathrm{t}$
$\mathrm{v}_{\mathrm{y}}=0+\frac{\mathrm{qE}_{0}}{\mathrm{~m}} \mathrm{t}$
$\left(3 \mathrm{v}_{0}\right)=\sqrt{\mathrm{v}_{\mathrm{y}}^{2}+\mathrm{v}_{0}^{2}}$
$\Rightarrow \mathrm{v}_{\mathrm{y}}^{2}=8 \mathrm{v}_{0}^{2}$
$\Rightarrow \frac{\mathrm{q} \mathrm{E}_{0}}{\mathrm{~m}} \mathrm{t}=2 \sqrt{2} \mathrm{v}_{0}$
$\Rightarrow \mathrm{t}=\frac{2 \sqrt{2} \mathrm{~m}}{\mathrm{q} \mathrm{E}_{0}} \mathrm{v}_{0}$
$\mathrm{t} \propto \frac{\mathrm{m}}{\mathrm{q}}$
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