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(B) The de-Broglie wavelength $\lambda$ of an electron accelerated by a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqr

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become $\frac{1}{2}$ times

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(B) The de-Broglie wavelength $\lambda$ of an electron accelerated by a potential difference $V$ is given by the formula: $\lambda = \frac{12.27}{\sqrt{V}} \mathring{A}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Given initial voltage $V_1 = 25 \ kV$ and final voltage $V_2 = 100 \ kV$.Let $\lambda_1$ be the initial wavelength and $\lambda_2$ be the final wavelength.$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = \sqrt{\frac{25}{100}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the new wavelength $\lambda_2 = \frac{1}{2} \lambda_1$.

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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