The work function of a certain metal is 3.31 | vedclass

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(B) The work function $W_0 = 3.31 	imes 10^{-19} \, J$.The wavelength of incident radiation is $\lambda = 5000 \, 	ext{Å} = 5000 	imes 10^{-10} \,

Vedclass · Accepted Answer

$0.41$

Vedclass · Answer

(B) The work function $W_0 = 3.31 	imes 10^{-19} \, J$.The wavelength of incident radiation is $\lambda = 5000 \, 	ext{Å} = 5000 	imes 10^{-10} \, m$.According to Einstein's photoelectric equation, the energy of the incident photon $E$ is given by $E = W_0 + KE_{max}$.First, calculate the energy of the incident photon:$E = \frac{hc}{\lambda} = \frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{5000 	imes 10^{-10}} = \frac{19.86 	imes 10^{-26}}{5 	imes 10^{-7}} = 3.972 	imes 10^{-19} \, J$.Now, calculate the maximum kinetic energy $KE_{max}$:$KE_{max} = E - W_0 = 3.972 	imes 10^{-19} \, J - 3.31 	imes 10^{-19} \, J = 0.662 	imes 10^{-19} \, J$.To convert this energy into electron-volts $(eV)$:$KE_{max} (eV) = \frac{0.662 	imes 10^{-19} \, J}{1.6 	imes 10^{-19} \, J/eV} = 0.41375 \, eV \approx 0.41 \, eV$.

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