Question

If the cathode of a photocell is illuminated with a light of increasing frequency, the anode current will start at a frequency of $3\times {10}^{14}Hz$. Now a capacitor of capacitance $1pF$ is connected between the anode and the cathode of this photocell and the cathode is illuminated with light of frequency $7\times {10}^{14}Hz$. Assuming the illumination is long enough, find the approximate number of electrons arriving the anode? ($h=6.63\times {10}^{-34}Js,e=1.6\times {10}^{-19}C)$

• A. $1\times {10}^7$
• B. $1\times {10}^8$
• C. $4\times {10}^8$
• D. $4\times {10}^7$

Answer 1

The correct option is A $1\times {10}^7$

From the photo electric equation,
$\Rightarrow hv=h{v}_0+eV.......\left(i\right)$
For the capacitor $V=\frac{q}{C}=\frac{ne}{C}....\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow hv=h{v}_0+e\frac{ne}{C}\Rightarrow n=\frac{h(v-v_0)C}{e^2}=1.03\times {10}^7$