The de Broglie wavelength of a charged partic | vedclass

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(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{

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$\frac{10 \lambda}{11}$

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(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqrt{2mqV}}$From this expression,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Given that the potential difference is increased by $21 \%$,the new potential difference $V^{\prime}$ is:$V^{\prime} = V + 0.21V = 1.21V$Let the new de Broglie wavelength be $\lambda^{\prime}$. Then:$\frac{\lambda^{\prime}}{\lambda} = \sqrt{\frac{V}{V^{\prime}}} = \sqrt{\frac{V}{1.21V}} = \sqrt{\frac{1}{1.21}} = \frac{1}{1.1} = \frac{10}{11}$Therefore,$\lambda^{\prime} = \frac{10}{11} \lambda$.

The de Broglie wavelength of a charged particle accelerated through a potential difference $V$ is $\lambda$. If the potential difference is increased by $21 \%$,the de Broglie wavelength of the charged particle is

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