Question

Photons of energy 6 eV are incident on a potassium surface whose work function is 2.1 eV. The corresponding stopping potential required is :

• A. 8.1 V
• B. 6 V
• C. 2. 1 V
• D. 3.9 V

Answer 1

The correct option is D 3.9 V

From photo-electric equation, $e{V}_0=E-\phi$

$e{V}_0=(6-2.1)eV$

$V_0=3.9V$

So, the answer is option (D).