Question

2. The threshold energy for photoelectric emission of electrons from a metal is 2.0eV. If light of energy 5.5 x 10-19 J falls on its surface, will the electrons be ejected or not? If yes, also calculate kinetic energy of emitted electrons. 3 What is the​

Answer 1

Answer:

Energy of incident radiation is E=

λ

hc

=

3×10

−7

6.6×10

−34

×3×10

8

=6.6×10

−19

J.

Also, work function is ϕ=2×1.6×10

−19

J=3.2×10

−19

J.

Hence, maximum kinetic energy of emitted photo electrons will be 6.6×10

−19

J−3.2×10

−19

J=3.4×10

−19

J.

Answer 2

Answer:

The kinetic energy of emmited electrons will be 6.6×10^-19.

Explanation:

Hope this will help.