Question

In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required, what is the work function of the metal?

Answer 1

Answer:

The  work function is  $\phi = 2.46 \ eV$

Explanation:

From the question we are told that

    The light energy is  $E = 3.56 eV$

     The  stopping voltage is  $V = 1.10 \ V$

Generally work function is mathematically represented as

      $\phi = E - KE$

Where KE is the kinetic energy of the ejected electron and it is mathematically represented as

         $KE = V * e$

Where  e is the charge on the electron

So  

        $KE = 1.10eV$

Thus  

        $\phi = 3.56eV - 1.10 eV$

=>      $\phi = 2.46 \ eV$