Question

1.
A photon of wavelength 5000 Å strikes a metal surface having
work function of 2.20 eV. The kinetic energy of the emitted
photo electron is
La) 4.4 x 10-20 J (b) 0.425 eV
(c) 2.20 eV
(d) No electron will be emitted​

Answer 1

Answer:

We know λ=5×10

–7

m(given)

C=3×10

8

From the equation E=hv orhc/λ

Where, h = Planck’s constant =6.626×10

–34

Js

c = velocity of light in vacuum =3×10

8

m/s

λ= wavelength of photon =5×10

–7

m

Substituting the values in the given expression of E:

E=(6.26×10

−34

)(3×10

8

)/5×10

−7

=3.9695×10

−19

J

Hence, the energy of the photon is 3.97×10

–19

J.

(ii) The kinetic energy of emission Ek is given by

=hv−hv

0

(E−W)eV

1.6020×10

−19

3.9695×10

−19

eV−2.20eV

=(2.481–2.20)eV

=0.281eV

1eV=1.6×10

−

19J

so 0.281eV=0.281×1.6×10

−19

=4.48×10

−20