Question

In photoelectric emission process from a metal of work function $1.8\, eV$, the kinetic energy of most energetic electrons is $0.5\, eV$. The corresponding stopping potential is

• A. 1.8V
• B. 1.3V
• C. 0.5 V
• D. 2.3 V

Answer 1

The Correct Option is C

$K .E _{\cdot \max }= eV _{0}$
$V _{0}=\frac{ K \cdot E _{\max }}{ e }=\frac{0.5 \,eV }{ e }$
$V _{0}=0.5$ volt

Answer 2

The stopping potential Vs is related to the maximum kinetic energy of the emitted electrons Kmax = e Vs Kmax =0.5 eV or Vs = 0.5 V

Therefore, the correct option is ‘C’ i.e 0.5V

Answer 3

In a photoelectric effect, the total maximum kinetic energy of an electron is, 

K_max= eV₀

So, V₀= \(\frac{k_{max}}{e}\) = \(\frac{0.5eV}{e}\) = 0.5V (as eV = 1 electronic charge time at 1 Volt)

Therefore, V₀= 0.5V