The stopping potential for photoelectrons emi | vedclass

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(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{6630 	imes 10^{-10}} = \frac{19.89 	imes 10^{-26}}{6.63 	imes 10^{-7}} = 3 	imes 10^{-19} \; J$.Converting to electron volts: $E = \frac{3 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 1.875 \; eV$.According to Einstein's photoelectric equation: $E = \phi + eV_{0}$,where $\phi$ is the work function and $V_{0}$ is the stopping potential.Given $V_{0} = 0.42 \; V$,so $eV_{0} = 0.42 \; eV$.Thus,$\phi = E - eV_{0} = 1.875 \; eV - 0.42 \; eV = 1.455 \; eV$.Converting work function to Joules: $\phi = 1.455 	imes 1.6 	imes 10^{-19} \; J = 2.328 	imes 10^{-19} \; J$.The threshold frequency $
u_{0}$ is given by $\phi = h
u_{0}$.$
u_{0} = \frac{\phi}{h} = \frac{2.328 	imes 10^{-19}}{6.63 	imes 10^{-34}} \approx 0.351 	imes 10^{15} \; s^{-1} = 35.1 	imes 10^{13} \; s^{-1}$.Rounding to the nearest integer,$x = 35$.

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