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(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + \frac{1}{2}mv_{\max}^2$. Let the initial wavelength be $\lambda_1 = \l

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$< \frac{1}{2}$

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(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + \frac{1}{2}mv_{\max}^2$. Let the initial wavelength be $\lambda_1 = \lambda$ and the final wavelength be $\lambda_2 = 4\lambda$. The maximum kinetic energy is given by $K_{\max} = \frac{hc}{\lambda} - W_0$. Since $K_{\max} = \frac{1}{2}mv_{\max}^2$,we have $v_{\max} = \sqrt{\frac{2}{m} (\frac{hc}{\lambda} - W_0)}$. When $\lambda$ increases to $4\lambda$,the energy of the incident photon $\frac{hc}{4\lambda}$ decreases. Since $K_{\max}$ decreases,$v_{\max}$ also decreases. Let the new velocity be $v'$. Then $v' = \sqrt{\frac{2}{m} (\frac{hc}{4\lambda} - W_0)}$. Comparing $v'$ with $v_{\max}$: $v' = \sqrt{\frac{1}{4} \frac{2hc}{\lambda} - \frac{2W_0}{m}} = \sqrt{\frac{1}{4} v_{\max}^2 - \frac{3W_0}{2m}}$. Since $\frac{1}{4} v_{\max}^2 - \frac{3W_0}{2m} Thus,the factor by which the velocity changes is less than $\frac{1}{2}$.

Assuming photoemission takes place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is

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