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(B) The energy of the incident photon is given by $E = \frac{12400}{\lambda (	ext{in } Å)} \, eV$.Given $\lambda = 1240 \, Å$, so $E = \frac{12400}{1

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$2 \, eV, 6200 \, Å$

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(B) The energy of the incident photon is given by $E = \frac{12400}{\lambda (	ext{in } Å)} \, eV$.Given $\lambda = 1240 \, Å$, so $E = \frac{12400}{1240} = 10 \, eV$.According to Einstein's photoelectric equation, $E = W + K_{max}$, where $K_{max} = e V_s$.Given stopping potential $V_s = 8 \, V$, so $K_{max} = 8 \, eV$.Therefore, the work function $W = E - K_{max} = 10 \, eV - 8 \, eV = 2 \, eV$.The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{12400}{W (	ext{in } eV)} \, Å$.$\lambda_0 = \frac{12400}{2} = 6200 \, Å$.

If a photocell is illuminated with a radiation of $1240 \, Å$, the stopping potential is found to be $8 \, V$; then the work function of the emitter and the threshold wavelength are

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