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If the stationary proton and $\alpha$ - particle are accelerated through same potential difference, the ratio of de-Broglie wavelength will be
Option: 1
$2$

Option: 2
$1$

Option: 3
$2 \sqrt{2}$

Option: 4
None of these

Answers (1)

best_answer

The gain in K.E. of a change particle after moving through a potential difference of $\mathrm{V}$ is given as $\mathrm{eV}$ , that is also equal to $\frac{1}{2} {mv}^2$ where $v$ is the velocity of the charge particle.

$\frac{1}{2} m v^2=q v$ $\Rightarrow$ $v=\sqrt{\frac{2 q V}{m}}$

$\therefore \quad \mathrm{mv}=\sqrt{2 \mathrm{mqV}}$

$\Rightarrow \quad \text { de-Broglie wavelength }=\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$

$\therefore \quad \frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha q_\alpha V_\alpha}{m_p q_p V_p}}$

$Putting\, \, \mathrm{V}_\alpha=\mathrm{V}_{\mathrm{p}}, \frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\sqrt{\frac{(4)(2)}{(1)(1)}}=2 \sqrt{2}$

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Ritika Kankaria

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