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प्रश्न
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
विकल्प
2 : 3
4 : 3
3: 4
3: 2
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उत्तर
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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संबंधित प्रश्न
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- The velocity of a photon is c.
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- In a photon-electron collision, both total energy and total momentum are conserved.
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Choose the correct answer from the options given below:
Einstein proposed that light energy is absorbed in discrete packets called what?
What condition must be met for photoelectric emission to occur?
How can maximum kinetic energy be written in terms of stopping potential \[V_0\]?
Why does the maximum kinetic energy of photoelectrons increase when frequency increases?
