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(B) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Initially,$\lambda = \frac{h}{mv_{0}}$.Finally,the wav

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$\frac{1}{E_{0}}$

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(B) The de-Broglie wavelength of an electron is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Initially,$\lambda = \frac{h}{mv_{0}}$.Finally,the wavelength becomes $\lambda' = \frac{\lambda}{3} = \frac{h}{mv'}$.Comparing the two,we get $\frac{\lambda}{\lambda/3} = \frac{h/mv_{0}}{h/mv'} \implies 3 = \frac{v'}{v_{0}}$,so the final velocity magnitude is $v' = 3v_{0}$.The electron experiences an acceleration $a = \frac{eE_{0}}{m}$ in the $y$-direction due to the electric field.The velocity vector after time $t$ is $\vec{v}' = v_{0}\hat{i} + \frac{eE_{0}t}{m}\hat{j}$.The magnitude of the final velocity is $|v'| = \sqrt{v_{0}^{2} + (\frac{eE_{0}t}{m})^{2}}$.Equating $|v'| = 3v_{0}$,we have $9v_{0}^{2} = v_{0}^{2} + (\frac{eE_{0}t}{m})^{2}$.This simplifies to $8v_{0}^{2} = (\frac{eE_{0}t}{m})^{2}$,which gives $t = \frac{m}{eE_{0}} \sqrt{8v_{0}^{2}}$.Thus,$t \propto \frac{1}{E_{0}}$.

An electron in an electron microscope with initial velocity $v_{0} \hat{i}$ enters a region of a stray transverse electric field $E_{0} \hat{j}$. The time taken for the change in its de-Broglie wavelength from the initial value of $\lambda$ to $\lambda / 3$ is proportional to

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