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(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.Th

Vedclass · Accepted Answer

$4$

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(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.This implies $\lambda \propto \frac{1}{\sqrt{V}}$.Let $\lambda_1$ be the wavelength at potential $V_1$,so $\lambda_1 = \frac{k}{\sqrt{V_1}}$,where $k$ is a constant.When the potential is changed to $V_2$,the wavelength increases by $50\%$,so $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1$.Thus,$\lambda_2 = \frac{k}{\sqrt{V_2}}$.Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{\sqrt{V_1}}{\sqrt{V_2}} = 1.5$.Squaring both sides: $\frac{V_1}{V_2} = (1.5)^2 = 2.25$.We can write $2.25$ as $\frac{225}{100} = \frac{9}{4}$.Given $(V_1/V_2) = (9/\alpha)$,comparing the two expressions gives $\alpha = 4$.

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