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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10}} \ J$.Converting to $eV$: $E = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10} 	imes 1.6 	imes 10^{-19}} \ eV \approx 3.09 \ eV$.According to Einstein's photoelectric equation: $E = W_0 + K_{max}$.Given the stopping potential $V_s = 2 \ V$,the maximum kinetic energy is $K_{max} = e \cdot V_s = 2 \ eV$.Therefore,the work function $W_0 = E - K_{max} = 3.09 \ eV - 2 \ eV = 1.09 \ eV$.Rounding to the nearest option,$W_0 \approx 1.1 \ eV$.

Light of wavelength $4000 \ \mathring A$ is incident on a photosensitive surface. If a potential of $-2 \ V$ is required to stop the emitted electrons,the work function of the material is: $(h = 6.6 \times 10^{-34} \ J \cdot s, e = 1.6 \times 10^{-19} \ C, c = 3 \times 10^8 \ m/s)$ (in $eV$)

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