Light source having wavelength 331 text{ nm} | vedclass

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the given values: $E = \frac{6.62 	imes 10^{-34} 	imes 3 	i

Vedclass · Accepted Answer

$5$

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the given values: $E = \frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{331 	imes 10^{-9}} = \frac{19.86 	imes 10^{-26}}{331 	imes 10^{-9}} = 0.06 	imes 10^{-17} 	ext{ J} = 6 	imes 10^{-19} 	ext{ J}$.According to Einstein's photoelectric equation, $E = \phi + K_{max}$, where $\phi$ is the work function and $K_{max}$ is the maximum kinetic energy.The maximum kinetic energy is $K_{max} = eV_s = 1.6 	imes 10^{-19} 	imes 0.2 = 0.32 	imes 10^{-19} 	ext{ J}$.Therefore, the work function $\phi = E - K_{max} = 6 	imes 10^{-19} - 0.32 	imes 10^{-19} = 5.68 	imes 10^{-19} 	ext{ J}$.Comparing this with $\alpha 	imes 10^{-19} 	ext{ J}$, we get $\alpha = 5.68$.

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