Question

The minimum energy required to overcome the attractive forces between electron and the surface of Ag metal is 7.52×10−19J.What will be the maximum kinetic energy of electron ejected out from Ag.Which is being exposed to uv light of λ=360A∘?

Answer 1

Answer:

Energy absorbed =hc/λ

=

360×10

−8

6.625×10

−27

×3×10

8

=5.52×10

−11

erg

or =5.52×10

−18

J [1erg=10

−7

J]

E

absorbed

= E used in attractive force + kinetic energy of the electron

=5.52×10

−18

J−5.52×10

−19

J

=49.68×10

−19

J