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(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda =

Vedclass · Accepted Answer

$14.14$

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(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For particle $A$: $\lambda_A = \frac{h}{\sqrt{2mq(50)}}$.For particle $B$: $\lambda_B = \frac{h}{\sqrt{2(4m)q(2500)}} = \frac{h}{\sqrt{8mq(2500)}} = \frac{h}{\sqrt{20000mq}}$.The ratio $\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{2(4m)q(2500)}}{\sqrt{2mq(50)}} = \sqrt{\frac{8m \cdot q \cdot 2500}{2m \cdot q \cdot 50}} = \sqrt{\frac{20000}{100}} = \sqrt{200} = 10\sqrt{2}$.Since $\sqrt{2} \approx 1.414$,the ratio is $10 	imes 1.414 = 14.14$.

$A$ particle $A$ of mass $m$ and charge $q$ is accelerated by a potential difference of $50 \ V$. Another particle $B$ of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500 \ V$. The ratio of de-Broglie wavelength $\frac{\lambda_A}{\lambda_B}$ is close to

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