Question

Ratio of frequencies of two
incident radiations on the same
metal is 1:2 and the ratio of
kinetic energy of photo
electrons is 2:5. Work function
of the metal is
(A) 1 eV
(B) 0.33 eV
(C) 0.66 eV
(D) 0.75 eV​

Answer 1

Answer:

some data is missing kindly check the question

Answer 2

BE = 2KEe

Assuming the KE value is 0.33 eV, work function of the metal = 0.66 eV

Explanation:

KEe = hf − BE

BE is the work function of the metal

KEe is the maximum kinetic energy of the ejected electron

hf is the photon's energy

Therefore BE = hf - KEe

Ratio of frequencies of two incident radiations on the same metal = 1:2

Ratio of KE of photo electrons = 2:5

Work function of the metal is the same for both radiations, BE.

Let f and 2f be the frequencies and 2KEe and 5KEe be the kinetic energies.

Then BE = hf1 - KE1 = hf2 - KE2

h(f) - 2KEe = h(2f) - 5KEe

5KEe - 2KEe = 2hf - hf

3KEe = hf

So the work function of the metal is BE = hf - KEe = 3KEe - KEe = 2KEe

The KE value is not mentioned in the question.

Assuming the KE value is 0.33 eV, then work function of the metal will be 2KEe = 0.66 eV.

Option C is the answer.