When light of a given wavelength is incident on a metallic surface, the minimum potential needed to

Question

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V . This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take, $\frac{h c}{e}=1.24 	imes 10^{-6} \mathrm{~J} \mathrm{~m} \mathrm{C}^{-1}$.]

Professor PC Thomas Classes · Answer

(a) : $V_{o_1}=6.0 \mathrm{~V}, V_{o_2}=0.6 \mathrm{~V} ; \lambda=\lambda, \lambda_2=4 \lambda$
$
I_1=I, I_2=\frac{I_1}{2}=\frac{I}{2} ; \frac{h c}{e}=1.24 	imes 10^{-6} \mathrm{Jm} \mathrm{C}^{-1}
$
Use the Einstein equation of photoelectric effect. \ Let $\phi$ be the work function
$
\begin{aligned}
& \frac{h c}{\lambda}=\phi+6 \
& \frac{h c}{4 \lambda}=\phi+0.6
\end{aligned}
$
By solving (i) and (ii), $\phi=1.2 \mathrm{eV}$ \ and $\lambda=1.72 	imes 10^{-7} \mathrm{~m}$