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  • The kinetic energy of an electron, and a proton are given as <img alt="4 \m

The kinetic energy of an electron, \alpha-\text{particle} and a proton are given as 4 \mathrm{~K}, 2 \mathrm{~K} \text{ and }\mathrm{K} respectively. The de-Broglie wavelength associated with electron ( \lambda \mathrm{e}), \alpha-particle ( \lambda \alpha) and the proton ( \lambda \mathrm{p})  are as follows :
 

Option: 1

\lambda \alpha>\lambda p>\lambda e


Option: 2

\lambda \alpha=\lambda p>\lambda e


Option: 3

\lambda \alpha=\lambda p<\lambda e


Option: 4

\lambda \alpha<\lambda p<\lambda e


Answers (1)

best_answer

According to De-Broglie, Momentum P=\frac{h}{\lambda}, where h is plank's constant and \lambda is wavelength.
Also, relation between Kinetic energy (\mathrm{KE}) and momentum (\mathrm{P})is given by: \mathrm{P}=\sqrt{2 \mathrm{mKE}}
Now, \lambda=\frac{h}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKE}}}
\begin{aligned} & \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{KE}_{\mathrm{e}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \times 4 \mathrm{k}}}=\frac{\mathrm{h}}{\sqrt{8 \mathrm{~m}_{\mathrm{e}} \mathrm{K}}} \\ & \lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{KE}_{\mathrm{p}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{k}}} \\ & \lambda_\alpha=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{\alpha}} \mathrm{KE}_\alpha}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{\alpha}} \cdot 4 \mathrm{k}}}=\frac{\mathrm{h}}{\sqrt{2 \times 2 \mathrm{~m}_{\mathrm{p} \cdot} \cdot \mathrm{k}}}=\frac{\mathrm{h}}{\sqrt{8 \mathrm{~m}_{\mathrm{p} \mathrm{K}}}} \end{aligned}
From the above data,

\lambda_\alpha<\lambda_p<\lambda_e

Posted by

Kuldeep Maurya

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