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(D) 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

Wavelength is decreased to $1/\sqrt{2}$ times.

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(D) 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 potential difference be $V_1 = V$ and the initial wavelength be $\lambda_1$.When the potential difference is doubled,the new potential difference is $V_2 = 2V$.The new wavelength $\lambda_2$ is given by $\lambda_2 = \frac{h}{\sqrt{2me(2V)}} = \frac{1}{\sqrt{2}} 	imes \frac{h}{\sqrt{2meV}}$.Therefore,$\lambda_2 = \frac{1}{\sqrt{2}} \lambda_1$.Thus,the de-Broglie wavelength changes by a factor of $1/\sqrt{2}$.

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

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