Question

The threshhold frequency for metal is 8×10^14 what is the kinetic energy of an electron emitted having frequency 1.0×10^15

Answer 1

Answer:

The kinetic energy is 1.324*10^-19 J

Explanation:

Given: v1 (emmited frequency) =1.0*10^15 and v2 (threshold frequency)= 8*10^14

Solution:

hv1=hv2+K.E.

K.E.= hv1-hv2

K.E.= h(v1-v2)

K.E.= 6.62*10^-34 (10*10^14- 8*10^14)

K.E.= 6.62*10^-34 (2*10^14)

K.E.= 6.62*10^-34 *2*10^14

K.E.= 1.324*10^-19 J

Hope it helps :)