Two light waves of wavelengths 600 ,nm and 20 | vedclass

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Question

(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = \frac{1}{2}mv^2$. For $\lambda_1 = 200 \,nm$,let the veloci

Vedclass · Accepted Answer

$\frac{hc}{8} 	imes 10^7 \,J$

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(A) According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi = \frac{1}{2}mv^2$. For $\lambda_1 = 200 \,nm$,let the velocity be $v$. Then $\frac{hc}{200 	imes 10^{-9}} - \phi = \frac{1}{2}mv^2$ $(i)$. For $\lambda_2 = 600 \,nm$,the velocity is $\frac{v}{3}$. Then $\frac{hc}{600 	imes 10^{-9}} - \phi = \frac{1}{2}m(\frac{v}{3})^2 = \frac{1}{9}(\frac{1}{2}mv^2)$ (ii). From $(i)$,$\frac{1}{2}mv^2 = \frac{hc}{200 	imes 10^{-9}} - \phi$. Substitute this into (ii): $\frac{hc}{600 	imes 10^{-9}} - \phi = \frac{1}{9}(\frac{hc}{200 	imes 10^{-9}} - \phi)$. Multiply by $9$: $\frac{9hc}{600 	imes 10^{-9}} - 9\phi = \frac{hc}{200 	imes 10^{-9}} - \phi$. $8\phi = hc(\frac{9}{600 	imes 10^{-9}} - \frac{1}{200 	imes 10^{-9}}) = hc(\frac{9-3}{600 	imes 10^{-9}}) = hc(\frac{6}{600 	imes 10^{-9}}) = hc(\frac{1}{100 	imes 10^{-9}}) = hc 	imes 10^7$. Therefore,$\phi = \frac{hc}{8} 	imes 10^7 \,J$.

Two light waves of wavelengths $600 \,nm$ and $200 \,nm$ are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is $\frac{1}{3}$ of the maximum velocity of the photoelectrons produced due to the other wavelength. The work function of the metal is:

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