Question

The wavelength and intensity of the incident light is 4000 A nad 0.1 W respectively.

What is the minimum change in the light energy​

Answer 1

Answer:

I don't know the exact,

Let number of photons emitted per second be n.

Then intensity =100W/m

2

λ

nhc

=100

n=

hc

100×λ

=

6.6×10

−34

×3×10

8

100×4000×10

−10

=2.02×10

20

Answer 2

Answer:

We know that,

We know that,Maximum energy of emitted photons electron KE energy of the incident radiation - Work function = 12.37eV

Now the kinetic energy is,

KE = 12.37 - 2

= 10.37eV

The kinetic energy is 10.37ev.