Question

Photons of frequency v are incident on the surfaces of two metals A and B
of threshold frequencies 3/4 v and 2/3 v, respectively. The ratio of maximum kinetic energy of electrons emitted from A to that from B is
(a) 2:3
(b) 4:3
(c) 3:4
(d) 3:2​

Answer 1

Answer:

3.4. h

shayad

if correct plzz like

Answer 2

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

Given :

Threshold frequency of metals A and B are 3/4 v and 2/3 v, respectively.

We know , kinetic energy of an electrons emitted is given by :

$K.E=h\nu-h\nu_o$

Here , $\nu\ and\ \nu_o$ are frequency of incident photons and threshold frequency respectively .

Therefore , ration of kinetic energy of electrons emitted from A to that from B is :

$\dfrac{K.E_A}{K.E_B}=\dfrac{h\nu-h\dfrac{3\nu}{4}}{h\nu-h\dfrac{2\nu}{3}}\\\\\dfrac{K.E_A}{K.E_B}=\dfrac{\dfrac{h\nu}{4}}{\dfrac{h\nu}{3}}\\\\\\\dfrac{K.E_A}{K.E_B}=\dfrac{3}{4}$

Therefore , potion (c) is correct .

Hence , this is the required solution .

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Quantum Physics

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