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

(d)1:3

(c) 3:4

(b) √3:√2

(NOTE :- Only genuine answer otherwise I will report you.)​

Answer 1

Answer:

2:3

Explanation:

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Answer 2

Answer:

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_oK.E=hν−hνo  { Here \nu_oνo is threshold frequency }

So , Kinetic energy of object A , K.E_A=h\nu-h\dfrac{\nu}{2}=\dfrac{h\nu}{2}K.EA=hν−h2ν=2hν .

Kinetic energy of object A , K.E_B=h\nu-h\dfrac{\nu}{3}=\dfrac{2h\nu}{3}K.EB=hν−h3ν=32hν .

Dividing them we get there ration as :

3 : 4 .

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