An electron,accelerated by a potential differ | vedclass

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Question

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula $\lambda = \frac{h}{\sqr

Vedclass · Accepted Answer

$2$

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(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula $\lambda = \frac{h}{\sqrt{2meV}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Let the initial wavelength be $\lambda = \frac{k}{\sqrt{V}}$ and the new wavelength be $\lambda^{\prime} = \frac{k}{\sqrt{4V}}$,where $k$ is a constant.Dividing the two expressions,we get $\frac{\lambda^{\prime}}{\lambda} = \frac{\sqrt{V}}{\sqrt{4V}} = \frac{1}{2}$.Therefore,$\lambda^{\prime} = \frac{\lambda}{2}$.Comparing this with the given expression $\lambda^{\prime} = \frac{\lambda}{n}$,we find that $n = 2$.

An electron,accelerated by a potential difference $V$,has de Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $4V$,its de Broglie wavelength will become $\frac{\lambda}{n}$ where $n=$

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