Question
A proton and an a particle are accelerated through the same potential difference. Which one of the two has (i) greater de-Broglie wavelength, and (ii) less kinetic energy? Justify your answer.
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Answer
- $\lambda =\frac{\text{h}}{\text{p}} =\frac{\text{h}}{\sqrt{2\text{mqV}}}$
$\lambda_{proton} > \lambda_{\propto} - particle$
(Also, $\frac{\lambda_{proton}}{\lambda_{\alpha}} =2\sqrt{2}(\text{ or }\sqrt{8} ) $
- K.E. = q V
$(\text{K.}\text{E} )_{proton} < (\text{K.}\text{E} )_{\alpha - particle}$
$\text{Also},\frac{(\text{K.E.})\propto}{(\text{K.E.})_{\rho}} = 2 ) $.
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