If lambda_0 and lambda are the threshold wave | vedclass

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(C) According to Einstein's photoelectric equation,the kinetic energy of the ejected electron is given by:$E = W + K.E.$$\frac{hc}{\lambda} = \frac{hc

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$\sqrt{\frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)}$

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(C) According to Einstein's photoelectric equation,the kinetic energy of the ejected electron is given by:$E = W + K.E.$$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2$Rearranging for the kinetic energy term:$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$$\frac{1}{2}mv^2 = hc \left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) = hc \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)$Solving for velocity $v$:$v^2 = \frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)$$v = \sqrt{\frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)}$

If $\lambda_0$ and $\lambda$ are the threshold wavelength and the wavelength of incident light,respectively,the velocity of the photoelectron ejected from the metal surface is:

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