A beam of photons with an energy of 10.5 eV | vedclass

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(C) Given: Energy of incident photons $E = 10.5 \ eV$.Maximum velocity of photoelectrons $v = 1.6 	imes 10^6 \ m \ s^{-1}$.Mass of electron $m = 9

Vedclass · Accepted Answer

$3.3$

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(C) Given: Energy of incident photons $E = 10.5 \ eV$.Maximum velocity of photoelectrons $v = 1.6 	imes 10^6 \ m \ s^{-1}$.Mass of electron $m = 9 	imes 10^{-31} \ kg$.The maximum kinetic energy $K.E._{\max}$ is given by $\frac{1}{2}mv^2$.$K.E._{\max} = \frac{1}{2} 	imes (9 	imes 10^{-31}) 	imes (1.6 	imes 10^6)^2 = 0.5 	imes 9 	imes 10^{-31} 	imes 2.56 	imes 10^{12} = 11.52 	imes 10^{-19} \ J$.Converting this to electron-volts: $K.E._{\max} = \frac{11.52 	imes 10^{-19}}{1.6 	imes 10^{-19}} \ eV = 7.2 \ eV$.According to Einstein's photoelectric equation, $E = \phi + K.E._{\max}$, where $\phi$ is the work function.$\phi = E - K.E._{\max} = 10.5 \ eV - 7.2 \ eV = 3.3 \ eV$.

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