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(B) The de-Broglie wavelength $\lambda$ associated with an electron accelerated by a potential $V$ is given by the relation $\lambda = \frac{1.227}{\s

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decrease by $2$ times

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(B) The de-Broglie wavelength $\lambda$ associated with an electron accelerated by a potential $V$ is given by the relation $\lambda = \frac{1.227}{\sqrt{V}} \; nm$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Let $\lambda_1$ be the initial wavelength at $V_1 = 25 \; kV$ and $\lambda_2$ be the final wavelength at $V_2 = 100 \; kV$.Taking the ratio,we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{100 \; kV}{25 \; kV}} = \sqrt{4} = 2$.Therefore,$\lambda_2 = \frac{\lambda_1}{2}$.This means the de-Broglie wavelength decreases by $2$ times.

Electrons used in an electron microscope are accelerated by a voltage of $25 \; kV$. If the voltage is increased to $100 \; kV$,then the de-Broglie wavelength associated with the electrons would

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