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(B) The kinetic energy acquired by an electron accelerated through a potential difference $V$ is given by $K.E. = eV = \frac{1}{2}mu^2$, where $u$ is

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$\sqrt{2meV}$

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(B) The kinetic energy acquired by an electron accelerated through a potential difference $V$ is given by $K.E. = eV = \frac{1}{2}mu^2$, where $u$ is the velocity of the electron.Rearranging for velocity: $u = \sqrt{\frac{2eV}{m}}$.According to the de Broglie relation, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mu}$, which implies $\frac{h}{\lambda} = mu$.Substituting the expression for $u$ into the equation for $\frac{h}{\lambda}$:$\frac{h}{\lambda} = m 	imes \sqrt{\frac{2eV}{m}} = \sqrt{m^2 	imes \frac{2eV}{m}} = \sqrt{2meV}$.

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