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

Vedclass · Accepted Answer

$4$

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(B) 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 relation, we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Therefore, $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.Given that $\lambda_1 = \lambda$ and $\lambda_2 = 2\lambda$, we substitute these values:$\frac{\lambda}{2\lambda} = \sqrt{\frac{V_2}{V_1}}$.$\frac{1}{2} = \sqrt{\frac{V_2}{V_1}}$.Squaring both sides, we get: $\frac{1}{4} = \frac{V_2}{V_1}$.Thus, $\frac{V_1}{V_2} = 4$.

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

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