A metallic surface is first irradiated with infrared radiation and photoelectrons are emitted from the surface. The infrared radiation is replaced by ultraviolet radiation of the same intensity. What will be the change in the kinetic energy of the photoelectrons and the rate at which they are ejected?

The kinetic energy will increase by using ultraviolet radiation.

The rate at which photo electrons are ejected remains same.

Explanation:

KINETIC ENERGY:

According to Einstein's Photoelectric Equation:

Energy of Photon = Work Function of Metal + Kinetic Energy

E = ∅ + K.E

K.E = hc/λ - ∅

where,

h = Plank's Constant

c = speed of light

λ = wavelength of radiation

So, for a metal work function is constant. Hence, the kinetic energy of the electrons depends upon the wavelength. Kinetic energy of photo electrons is inversely proportional to the wavelength of the radiation.

The wavelength of infrared radiation is greater than the wavelength of ultraviolet.

Therefore, the kinetic energy will increase by using ultraviolet radiation.

RATE OF EJECTION:

The rate of ejection of electrons depends upon the intensity of the photons or radiation only. Since, the intensity of both radiations is same.

Therefore, the rate at which photo electrons are ejected remains same.

f is prop to m1xm2 double m1 double force

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