Question

Photo-energy 6 eVare incident on a surface of work function 2.1 eV. What are the stopping potential

• A. – 5V
• B. – 1.9 V
• C. – 3.9 V
• D. – 8.1 V

Answer 1

Answer: (C)

– 3.9 V

Answer 2

By using Einstein's equation E = W₀ + K_max

⇒ $6 = 2.1 + K _ { \max }$ ⇒ $K _ { \max } = 3.9 \mathrm { eV }$

Also $V _ { 0 } = - \frac { K _ { \max } } { \rho } = - 3.9 \mathrm {~V}$