Kinetic energy of an electron emitted when radiation of frequency v= 1.25 x10^15 strikes the metal surface whose threshold frequency v0 = 7.25 x 10^14 is

Asked by arushidabhade | 29th Jun, 2019, 08:42: PM

Expert Answer:

Given:
 
K.E. = 1.25 x1015 s-1
 
Threshold frequency ν0 = 7.25 x 1014 s-1
 
Plank's constant, h = 6.62 × 10-34 m2kg/s
 
Energy of radiation = hν0 = h(v-ν0)
 
                            = 6.62 × 10-34 (12.51014  - 7.25 x 1014)
 
                            =3.5 x 10-19 J

Answered by Varsha | 29th Jun, 2019, 11:16: PM