Question

The work function of a metal is 2.5 × 10−19 J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency 6.0 × 1014 Hz, what will be the stopping potential?

Answer 1

Explanation:

Given data in the question  

Metal work function,$W_{o}=2.5 \times 10^{-19} \mathrm{J}$

Light Beam frequency, v = 6.0 × 1014 Hz

(a) metal  Work function

  $W_{0}=h v_{0}$,

  where h is Planck's constant

               $v_{0}$ frequency threshold

$v_{0}=\frac{W_{0}}{h}$

$\mu_{0}=\frac{2.5 \times 10^{-19}}{6.63 \times 10^{-34}}$

$=3.8 \times 10^{14} \mathrm{Hz}$

(b) The Photoelectric Equation of Einstein,

where v = light frequency  

         $V_{0}$ =  potential Stopping  

         e = electron charge  

$V_{0}=\frac{h v-W_{0}}{E}$

$=\frac{6.63 \times 10^{-34} \times 6 \times 1014-2.5 \times 10^{-19}}{1.6 \times 10-19}$

$=\frac{3.97 \times 10^{-19}-2.5 \times 10^{-19}}{1.6 \times 10^{-19}}=0.91 V$