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(B) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.

Vedclass · Accepted Answer

$\frac{hc}{2}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.For wavelength $\lambda_1$,the kinetic energy is $K_1 = \frac{hc}{\lambda_1} - W_0$.For wavelength $\lambda_2$,the kinetic energy is $K_2 = \frac{hc}{\lambda_2} - W_0$.Given that $K_2 = 3K_1$,we substitute the expressions:$\frac{hc}{\lambda_2} - W_0 = 3\left(\frac{hc}{\lambda_1} - W_0\right)$.$\frac{hc}{\lambda_2} - W_0 = \frac{3hc}{\lambda_1} - 3W_0$.Rearranging the terms to solve for $W_0$:$3W_0 - W_0 = \frac{3hc}{\lambda_1} - \frac{hc}{\lambda_2}$.$2W_0 = hc\left(\frac{3}{\lambda_1} - \frac{1}{\lambda_2}\right)$.$2W_0 = hc\left(\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.Therefore,$W_0 = \frac{hc}{2}\left(\frac{3\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\right)$.

When the wavelength of incident radiation on a metal surface is reduced from $\lambda_1$ to $\lambda_2$,the kinetic energy of the emitted photoelectrons is tripled. Find the work function of the metal. [$h =$ Planck's constant,$c =$ velocity of light]

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