Question

The stopping potential, when a metal with work function 0.6 eV is illuminated with light of energy 2 eV will be :

• A. 1.4 V
• B. 2.8 eV
• C. 4.2 eV
• D. 0.7 V

Answer 1

The correct option is A 1.4 V

Work-function $=0.6eV$

and $E=2eV$

Hence; Using Photoelectric effect :

$E=work-function+(K.E{)}_{max}$

$=\varphi +(K.E.{)}_{max}$

Hence, $(K.E.{)}_{max}=E-\varphi$

$\Rightarrow e{V}_0=(2-0.6)eV$

$\Rightarrow V_0=1.4volts$