A copper ball of radius 1, cm and work functi | vedclass

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400}{\lambda(	ext{in } \mathring{A})} \, eV$.Substituting $\lambd

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$5.5 	imes 10^{-13}\, C$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400}{\lambda(	ext{in } \mathring{A})} \, eV$.Substituting $\lambda = 2500\, \mathring{A}$,we get $E = \frac{12400}{2500} = 4.96\, eV$.The maximum kinetic energy of the emitted electrons is $K_{max} = E - \Phi$,where $\Phi = 4.47\, eV$ is the work function.$K_{max} = 4.96 - 4.47 = 0.49\, eV$.As the ball emits electrons,it becomes positively charged,creating a potential $V$ that stops further emission. The stopping potential $V$ is equal to $K_{max}/e$,so $V = 0.49\, V$.The potential of a charged sphere is $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} = k \frac{Q}{r}$,where $k = 9 	imes 10^9\, N\cdot m^2/C^2$ and $r = 0.01\, m$.$0.49 = \frac{9 	imes 10^9 	imes Q}{0.01}$.$Q = \frac{0.49 	imes 0.01}{9 	imes 10^9} = \frac{0.0049}{9 	imes 10^9} \approx 5.44 	imes 10^{-13}\, C$.Rounding to the nearest option,$Q = 5.5 	imes 10^{-13}\, C$.

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