Question

If the stopping potential of a metal when illuminated with a radiation of wavelength 480 nm is 1.2 V, find (a) the work function of the metal, (b) the cutoff wavelength of the metal, and (c) the maximum energy of the ejected electrons

Answer 1

Answer:

a) $\phi=2.22\times10^{-19}\, J$

b) $\lambda_0=900 \, nm$

c) $K_{max}=e\cdot V_0=1.922\times 10^{-19}\, J$

Explanation:

a)  

We have that for the photoelectric effect the maximum kinetic energy for the electrons is given by:

$K_{max}=\frac{hc}{\lambda}-\phi=e\cdot V_0$

From the previous we get:

$\phi=\frac{hc}{\lambda}-e\cdotV_0\=2.22\times10^{-19} \, J$

Where we used the fact that $h=6.63\times 10^{-34} \, J.s$,$e=1.602\times 10^{-19} \, C$ and $c=3\times 10^{8}\, m/s$.

b)

The work function is defined as:

$\phi=\frac{hc}{\lambda_{0}}\implies\lambda_0=\frac{hc}{\phi}=9.0\times10^{-7}=900 \, nm$

Where $\lambda_0$ is the cutoff wavelength.

c)

As said before:

$K_{max}=e\cdotV_0=1.602\times 10^{-19}=1.922\times 10^{-19} J$