A silver sphere of radius 1 cm and work func | vedclass

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(A) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{1240 \ eV \cdot nm}{200 \ nm} = 6.2 \ eV$.The maximum kinetic energy of the e

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$7$

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(A) The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{1240 \ eV \cdot nm}{200 \ nm} = 6.2 \ eV$.The maximum kinetic energy of the emitted photoelectrons is $K_{max} = E - \phi = 6.2 \ eV - 4.7 \ eV = 1.5 \ eV$.As the sphere emits photoelectrons,it becomes positively charged and its potential $V$ increases. Emission stops when the potential $V$ is such that the energy required to remove an electron is equal to the maximum kinetic energy,i.e.,$eV = K_{max} = 1.5 \ eV$. Thus,$V = 1.5 \ V$.The potential of a sphere of radius $R$ with charge $q = ne$ is $V = \frac{kq}{R} = \frac{k(ne)}{R}$,where $n$ is the number of photoelectrons.Substituting the values: $1.5 = \frac{(9 	imes 10^9) 	imes n 	imes (1.6 	imes 10^{-19})}{10^{-2}}$.$1.5 = n 	imes 1.44 	imes 10^{-7} \implies n = \frac{1.5}{1.44} 	imes 10^7 \approx 1.04 	imes 10^7$.Given $n = A 	imes 10^Z$,we have $1.04 	imes 10^7 = A 	imes 10^Z$. Comparing,$Z = 7$.

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