For intensity I of a light of wavelength 5000 | vedclass

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(C) According to Einstein's photoelectric equation: $E = W_0 + K_{\max}$.Here,$E$ is the energy of the incident photon,$W_0$ is the work function,and

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$1.10$

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(C) According to Einstein's photoelectric equation: $E = W_0 + K_{\max}$.Here,$E$ is the energy of the incident photon,$W_0$ is the work function,and $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons.The energy of the incident photon is given by $E = \frac{12375}{\lambda(\mathring{A})} eV = \frac{12375}{5000} eV = 2.475 eV$.The maximum kinetic energy is related to the stopping potential $V_0$ by $K_{\max} = e V_0 = 1.36 eV$.Substituting these values into the equation: $2.475 eV = W_0 + 1.36 eV$.Therefore,$W_0 = 2.475 eV - 1.36 eV = 1.115 eV \approx 1.10 eV$.

For intensity $I$ of a light of wavelength $5000 \mathring{A}$,the photoelectron saturation current is $0.40 \mu A$ and the stopping potential is $1.36 V$. The work function of the metal is ............. $eV$.

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