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Question

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

Vedclass · Accepted Answer

Wavelength decreased to half

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(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the relation: $\lambda = \frac{h}{\sqrt{2meV}}$.This implies that $\lambda \propto \frac{1}{\sqrt{V}}$.Let the initial potential be $V_1$ and the final potential be $V_2 = 4V_1$.The ratio of the wavelengths is given by: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$.Substituting the value of $V_2$: $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{4V_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$\lambda_2 = \frac{\lambda_1}{2}$.Thus,the de-Broglie wavelength decreases to half of its initial value.

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

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