Question

The work function for the surface of aluminium is 4.2 eV. How much potential difference will be required to stop the emission of maximum energy electrons emitted by light of 2000 $\dot{A}$ wavelength? What will be the wavelength of that incident light for which stopping potential will be $0$ ?

Answer 1

Given, Work function$4.2eV$

$\lambda =$ wavelength $=200\mathring{A}$

When stopping potential will be $0$

the wavelength$=?$

Work function of the metal $Q_0=4.2eV=4.2\times 1.6\times {10}^{-19}J=6.72\times {10}^{-14}J$

wavelength of incident radiation

$\lambda =200\mathring{A}=2000\times {10}^{-10}m$

Now, using the formula for energy of a photon,

$E=\frac{hc}{\lambda }E=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2000\times {10}^{-10}}=9.93\times {10}^{-19}J$

As energy of incident photon

$E>Q_0$ hence photoelectric emission will take place.