Question

solve this one.......​

Answer 1

Answer:

Given :

frequency of light = f = 6 × 10^14 Hz

threshold frequency of light = fo = 1.2 × 10^14 Hz

h = 6.6 × 10^-34 J

e = 1.6 ×10^-19 C

To find : Cut - off potential = Vo = ?

Solution :

We have,

Vo = (h/e) × (f - fo)

= (6.6 × 10^-34/1.6 × 10^ -19) × (6 × 10^14 - 1.2×10^14)

= (4.125 × 10^-15) × (4.8 × 10^14)

= 19.8 × 10^-15 × 10^14

= 1.98 V

=> Vo = 1.98 V

Thus, the stopping potential of light waves is 1.98 V

Answer 2

Given :

frequency of light = f = 6*10^14 Hz

threshold frequency = fo = 1.2*10^14 Hz

h = 6.6*10^-34 J-s

e = 1.6*10^-19 C

To find : Stopping potential = V = ?

We have,

V = (h/e) * (f - fo)

V = (6.6*10^-34) * (6*10^14 - 1.2*10^14)

V = 1.98 V

Hence, the stopping potential is 1.98 V