Question

If photons of frequency v are incident on the surfaces of metals. A & B of
threshold frequencies v/2 and v/3 respectively, the ratio of the maximum
kinetic energy of electrons emitted from A to that from B is
(a) 2:3
(b) 3:4
(c) 1:3
(d) root 3:root2

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Answer 1

The ratio of the maximum  kinetic energy of electrons emitted from A to that from B is 3 : 4.

Given :

Frequency of photons , v .

Threshold frequency of metal A & B of  threshold frequencies v/2 and v/3 respectively.

We need to find the ratio of the maximum  kinetic energy of electrons emitted from A to that from B.

We know , K.E is given by :

$K.E=h\nu-h\nu_o$  { Here $\nu_o$ is threshold frequency }

So , Kinetic energy of object A , $K.E_A=h\nu-h\dfrac{\nu}{2}=\dfrac{h\nu}{2}$.

Kinetic energy of object A , $K.E_B=h\nu-h\dfrac{\nu}{3}=\dfrac{2h\nu}{3}$.

Dividing them we get there ration as :

3 : 4 .

Hence , this is the required solution.

Learn More :

Kinetic energy of electron

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