Question

In a photoelectric experiment with a monochromatic light I-V curve is as shown in figure. If work function of metal is 2 eV and electrons to photons ratio is ${10}^{-5}$, then [Take hc=1240 eV-nm]

• A. Power of light used is 70 W
• B. Number of photons incident on metal per second is $6.25\times {10}^{18}$
• C. Wavelength of light used is 177 nm
• D. If work function of metal decreases from 2 eV saturated current increases from $10\mu A$

Answer 1

Answer: (B), (C)

The correct options are
B Number of photons incident on metal per second is $6.25\times {10}^{18}$
C Wavelength of light used is 177 nm

By $I=\frac{N_e}{t}e⟹10\times {10}^{-6}=\frac{N_e\times 1.6\times {10}^{-19}}{1}{N}_e=\frac{10\times {10}^{13}}{1.6}$ $N_e$ is number of electrons per second.

For ${10}^5$ photons we get 1 electron so
$N_p={10}^5\times N_e=\frac{10\times {10}^{18}}{1.6}=\frac{{10}^{19}}{1.6}=6.25\times {10}^{18}$

Power of source is $P=N_p\frac{hc}{\lambda }$
Using $\frac{hc}{\lambda }-\varphi =e{V}_0$ is $\frac{hc}{\lambda }=5+2=7eV$
$P=\frac{{10}^{19}}{1.6}\times 7\times 1.6\times {10}^{-19}=7W$
By $\frac{hc}{\lambda }=7⟹\frac{1240eV-nm}{\lambda }=7eV$
$\lambda =\frac{1240}{7}nm=177nm$


As work function decreases the number of electrons leaving the metal will NOT change, only they will get ejected with higher energy. Hence, saturation current will not change.