An electron beam used in an electron microsco | vedclass

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Question

(D) 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

$\frac{\lambda_0}{\sqrt{2}}$

Vedclass · Answer

(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 expression,it is clear that $\lambda \propto \frac{1}{\sqrt{V}}$.Given the initial voltage $V_1 = 20\,kV$ and initial wavelength $\lambda_1 = \lambda_0$.The new voltage is $V_2 = 40\,kV$.Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$,we get:$\frac{\lambda_2}{\lambda_0} = \sqrt{\frac{20}{40}}$$\frac{\lambda_2}{\lambda_0} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$Therefore,the new wavelength is $\lambda_2 = \frac{\lambda_0}{\sqrt{2}}$.

An electron beam used in an electron microscope,when accelerated by a voltage of $20\,kV$,has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to $40\,kV$,then the de-Broglie wavelength associated with the electron beam would be:

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