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Question
Choose the correct answer from given options
Photons of frequency v are incident on the surface of two metals A and B of threshold frequency 3/4 v and 2/3 v, respectively. The ratio of maximum kinetic energy of electrons emitted from A to that from B is
Options
2 : 3
4 : 3
3: 4
3: 2
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Solution
According to Einstein's photoelectric equation,
hv = hv0 + Kmax
Where,
v = frequency of the incident light
v0 = threshold frequency of the metal
Kmax = maximum kinetic energy of the emitted photoelectrons
For the first metal,
`hv = h (3/4v)+ K_1`
⇒ `K_1 = (hv)/4`
For the second metal,
`hv = h(2/3v)+ K_2`
⇒ `K_2 = (hv)/3`
Thus, the ratio of the maximum kinetic energies is given as:
`K_1/K_2 = 3/4`
Hence, the correct answer is option 3 : 4.
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