Question

A vaccum photocell consists of a central cathode and an anode. The internal surface is silver of work function 4.5 eV. The contact potential difference between the electrodes equals to 0.6 V. The photocell is illuminated by light of wavelength $2.3\times {10}^{-7}m$.
a. What retarding potential difference should be applied between electrodes of the photocell for the photocurrent to drop to zero?
b. If a retarding potential of 1 V is applied between electrodes at what limiting wavelength $\lambda$ of light incident on the cathode will the photoelectric effect begin?

Answer 1

$\left(a\right)$ Contact potential difference;

$u=0.6V$

The Energy acquired by electron on realing the elected $=e{V}_0$

According to Eimstein;s Equation

$h\nu =W+KE-e{\mu }_0$

$\therefore (KE{)}_{max}=e{V}_0\to$ stopping potential

$e{V}_0=h\nu -W+3e{\mu }_0$

$\Rightarrow v_0=\frac{hc}{{\lambda }_e}-\frac{W}{e}+{\mu }_0$

$=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2.3\times {10}^{-7}\times {10}^{-7}\times 1.6\times {10}^{-19}}-4.5+0.6$

$=5.4-4.5+0.6$

$=1.5V$

$\left(b\right)$ Using above formula:-

$V_0=\frac{hc}{\lambda e}-\frac{W}{e}+40$

$1=\frac{hc}{\lambda e}-4.5+0.6$

$\Rightarrow \frac{hc}{\lambda e}=4.5-0.6-1$

$\Rightarrow \lambda =\frac{hc}{4.9e}=\frac{(6.62\times {10}^{-34})(3\times {10}^8)}{4.9\times 1.6\times {10}^{-19}}$

$=2540\times {10}^{-10}$

$=2540A^o$