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  • If   is the de Broglie wavelength for a proton accelerated through a potential difference of <img alt="

If \lambda_0  is the de Broglie wavelength for a proton accelerated through a potential difference of 100 \mathrm{~V}, the de Broglie wavelength for \alpha-particle accelerated through the same potential difference is:
 

Option: 1

2 \sqrt{2} \lambda_0


Option: 2

\frac{\lambda_0}{2}


Option: 3

\frac{\lambda_0}{2 \sqrt{2}}


Option: 4

\frac{\lambda_0}{\sqrt{2}}


Answers (1)

best_answer

Kinetic energy gained by a charge \mathrm{q} after being accelerated through a potential difference \mathrm{V} is given by

\mathrm{qV}=\frac{1}{2} \mathrm{mv}^2 ; \mathrm{v}=\sqrt{\frac{2 \mathrm{qV}}{\mathrm{m}}} \ and \ \mathrm{mv}=\sqrt{2 \mathrm{mqV}}

de Broglie wavelength, \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}

For a proton, \lambda_p=\frac{h}{\sqrt{2 m_p q_p V_p}}

For an \alpha-particle

\begin{aligned} & \lambda_\alpha=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_\alpha \mathrm{q}_\alpha \mathrm{V}_\alpha}} \quad \therefore \quad \frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}}}{\mathrm{m}_\alpha \mathrm{q}_\alpha \mathrm{V}_\alpha}} \\ & \mathrm{V}_\alpha=\mathrm{V}_{\mathrm{p}}=100 \mathrm{~V} \text { (Given) } \\ & \therefore \frac{\lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_\alpha \mathrm{q}_\alpha}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\left(4 \mathrm{~m}_{\mathrm{p}}\right)\left(2 \mathrm{q}_{\mathrm{p}}\right)}}=\sqrt{\frac{1}{8}}=\frac{1}{2 \sqrt{2}} \\ & \lambda_\alpha=\frac{\lambda_{\mathrm{p}}}{2 \sqrt{2}}=\frac{\lambda_0}{2 \sqrt{2}} \quad\left(\because \lambda_{\mathrm{p}}=\lambda_0(\text { Given) })\right. \end{aligned}

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Rishi

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