Question

Given that a photon of light of wavelength $10,000\stackrel{o}{A}$ has an energy equal to 1.23 eV. When light of wavelength $5000\stackrel{o}{A}$ and intensity $I_0$ falls on a photoelectric cell, the saturation current is $0.40\times {10}^{-6}A$ and the stopping potential is 1.36 V; then the work function is :

• A. 0.43 eV
• B. 1.10 eV
• C. 1.36 eV
• D. 2.47 eV

Answer 1

The correct option is B 1.10 eV

The Einstein's equation for photoelectric effect is given by $e{V}_0=\frac{hc}{\lambda }-W$ where $V_0=$ stopping potential, $h=$ Planck's constant, $c=$ velocity of light, $\lambda =$wavelength of incident light and $W=$ work function of metal.

Here given , $V_0=1.36V,\lambda =5000\times {10}^{-10}m$

So, $W=\frac{hc}{\lambda }-e{V}_0$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{5000\times {10}^{-10}}-(1.6\times {10}^{-19}\times 1.36)$

$=1.78\times {10}^{-19}J=1.10eV$ as $(1eV=1.6\times {10}^{-19}J)$