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(B) The maximum kinetic energy of the emitted photoelectrons is given by Einstein's photoelectric equation:$(KE)_{\max} = h
u - \phi = 4.2 	ext{ eV}

Vedclass · Accepted Answer

$1.25 	imes 10^8$

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(B) The maximum kinetic energy of the emitted photoelectrons is given by Einstein's photoelectric equation:$(KE)_{\max} = h
u - \phi = 4.2 	ext{ eV} - 2.4 	ext{ eV} = 1.8 	ext{ eV}$.Emission stops when the potential $V$ of the sphere reaches a value such that the potential energy of an electron equals its maximum kinetic energy:$eV = (KE)_{\max} \implies V = 1.8 	ext{ V}$.The potential of a metallic sphere of radius $r$ is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{q}{r}$.Substituting the values: $1.8 = (9 	imes 10^9) 	imes \frac{q}{0.1}$.Solving for charge $q$: $q = \frac{1.8 	imes 0.1}{9 	imes 10^9} = 2 	imes 10^{-11} 	ext{ C}$.The number of photoelectrons $n$ emitted is $n = \frac{q}{e} = \frac{2 	imes 10^{-11}}{1.6 	imes 10^{-19}} = 1.25 	imes 10^8$.

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