Question

If the momentum of a photon is 4 × 10^-21 kgm/s then its energy will be ?

Answer 1

Answer:

Energy of photon is given by

E=

λ

hc

............(i)

where h is the Planck's constant, c the velocity of light and λ its wavelength.

de-Broglie wavelength is given by

λ=

p

h

............(ii)

where p is being momentum of photon.

From Eqs. (i) and (ii), we get

E=

h/p

hc

=pc

or p=E/c

Given E=1 Mev=1×10

6

×1.6×10

−19

J,

c=3×10

8

m/s

Hence, after putting numerical values we obtain

p=

3×10

8

1×10

6

×1.6×10

−19

kg−m/s

=5×10

−22

kg−m/s

Please mark me as Braillest

Answer 2

Answer:

12×10^-13J this is the answer