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(D) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2 m_e e

Vedclass · Accepted Answer

$2500$

Vedclass · Answer

(D) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2 m_e e V}}$.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 have $\frac{\lambda}{2\lambda} = \sqrt{\frac{V_2}{10000}}$.Squaring both sides, we get $\frac{1}{4} = \frac{V_2}{10000}$.Thus, $V_2 = \frac{10000}{4} = 2500 \ V$.

An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential, so that the wavelength is doubled (in $V$)?

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