Question

A monochromatic light source of intensity 5 mW emits 8 × 1015 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

Answer 1

The work function of the metal is $1.906 e V$

Explanation:

Given data in the question  

light Intensity, I = 5 mW

Number of emitted photons per second, $n=8 \times 10^{15}$

potential ends,$V_{0}=2 \mathrm{V}$

Energy, $E=\mathrm{hv}=\frac{1}{\mathrm{n}}=\frac{5 \times 10^{-3}}{8 \times 10^{15}}$

Work function from Einstein's photoelectrical equation,

$W_{0}=h v-e V_{0}$

Here, h is Planck's constant

  $e=1.6 \times 10^{-19} \mathrm{C}$

When the respective values are replaced we get:

$W_{0}=\frac{5 \times 10^{-3}}{8 \times 10^{15}}-1.6 \times 10^{-19} \times 2$

     $=6.25 \times 10^{-19}-3.2 \times 10^{-19}$        

     $\begin{aligned}&=3.05 \times 10^{-19}\\&=\frac{3.05 \times 10^{-19}}{1.6 \times 10^{-15}}\end{aligned}$  

     $=1.906 e V$

Answer 2

Answer:

1.906 eV answer ..............