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

Vedclass · Accepted Answer

$\frac{\lambda}{3}$

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(B) 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 $\lambda_1 = \lambda$ at potential $V_1 = V$.Let the new wavelength be $\lambda_2$ at potential $V_2 = 9V$.Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$.Substituting the values: $\frac{\lambda_2}{\lambda} = \sqrt{\frac{V}{9V}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.Therefore,$\lambda_2 = \frac{\lambda}{3}$.

An electron accelerated by a potential difference $V$ has a de-Broglie wavelength $\lambda$. If the electron is accelerated by a potential difference $9V$,its de-Broglie wavelength will be

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