Light of wavelength 1824 mathring A,incident | vedclass

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(D) According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $E = \frac{hc}{\lambda}$ is the energy of the incident photon,$W_0$ is t

Vedclass · Accepted Answer

$1.5$

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(D) According to Einstein's photoelectric equation,$E = W_0 + K_{\max}$,where $E = \frac{hc}{\lambda}$ is the energy of the incident photon,$W_0$ is the work function,and $K_{\max}$ is the maximum kinetic energy.Using the relation $E \approx \frac{12400}{\lambda (	ext{in } \mathring A)} \ eV$:For $\lambda_1 = 1824 \ \mathring A$,$E_1 = \frac{12400}{1824} \approx 6.797 \ eV \approx 6.8 \ eV$.Given $K_{\max, 1} = 5.3 \ eV$,then $W_0 = E_1 - K_{\max, 1} = 6.8 - 5.3 = 1.5 \ eV$.For $\lambda_2 = 1216 \ \mathring A$,$E_2 = \frac{12400}{1216} \approx 10.197 \ eV \approx 10.2 \ eV$.Given $K_{\max, 2} = 8.7 \ eV$,then $W_0 = E_2 - K_{\max, 2} = 10.2 - 8.7 = 1.5 \ eV$.Thus,the work function of the metal surface is $1.5 \ eV$.

Light of wavelength $1824 \mathring A$,incident on the surface of a metal,produces photo-electrons with maximum energy $5.3 \ eV$. When light of wavelength $1216 \mathring A$ is used,the maximum energy of photoelectrons is $8.7 \ eV$. The work function of the metal surface is .............. $eV$.

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