Question

Anode voltage is at +3 V. Incident radiation has frequency 1.4 × 105 Hz and work function of photocathode is 2.8 eV. Find the minimum and maximum kinetic energy of photoelectrons in eV

Answer 1

Answer:

.........................

Answer 2

Explanation:

As we had provided a voltage , Kmin = +3

Kmax = 3+ (hv- Ф)

Note that frequency is 10^15 not 5

By solving , we get

Kmax = 6