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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $\lambda = 221 \,nm = 221 	imes 10^{-9} \,m$, $h = 6.63 	imes 10^{-

Vedclass · Accepted Answer

$1.49$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $\lambda = 221 \,nm = 221 	imes 10^{-9} \,m$, $h = 6.63 	imes 10^{-34} \,J \cdot s$, and $c = 3 	imes 10^8 \,m/s$.$E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{221 	imes 10^{-9}} \,J = \frac{19.89 	imes 10^{-26}}{221 	imes 10^{-9}} \,J = 0.09 	imes 10^{-17} \,J = 9 	imes 10^{-19} \,J$.Converting this energy into electron-volts $(eV)$:$E = \frac{9 	imes 10^{-19} \,J}{1.6 	imes 10^{-19} \,J/eV} = 5.625 \,eV \approx 5.63 \,eV$.The work function of the lead ball is $\phi = 4.14 \,eV$.The maximum kinetic energy of the emitted photoelectrons is $K_{max} = E - \phi = 5.63 \,eV - 4.14 \,eV = 1.49 \,eV$.As the ball loses electrons, it becomes positively charged and its potential increases until the photoelectrons can no longer escape. The maximum potential $V$ attained is given by $eV = K_{max}$.Thus, $V = 1.49 \,V$.

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