Question

For photoelectric emission from certain metal the cut off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass):
1. \(\sqrt{\dfrac{h\nu}{m}}\)
2. \(\sqrt{\dfrac{2h\nu}{m}}\)
3. \(2\sqrt{\dfrac{h\nu}{m}}\)
4. \(\sqrt{\dfrac{h\nu}{2m}}\)

Answer 1

Correct Answer - Option 2 : \(\sqrt{\dfrac{2h\nu}{m}}\)

CONCEPT:

• The work function is the minimum amount of energy required to release the electron from a photoemissive surface and is given by

⇒ ϕ = h ν0

Where h = Plancks constant, ν0 = threshold frequency

• The Ensitens photoelectric equation gives the kinetic energy of a photoelectron, the kinetic energy of a photoelectron is the difference in energy between the incident photon and work function of the material and is given by

⇒ Kinetic energy (KE) = hν - ϕ 

Where ν = frequency of incident light, h = Plancks constant, ϕ  = Work function

EXPLANATION: 

• The kinetic energy of photoelectron is given by

⇒ KE = hν - ϕ 

\(⇒ \frac{1}{2}mV^{2} = hν -\phi \) (For incident frequency ν )

• If the frequency is doubled then the above equation can be written as

\(⇒ \frac{1}{2}mV^{2} = 2hν -hν_{0}\)

Assume ν = ν₀, the above equation can be written as

\(⇒ \frac{1}{2}mV^{2} = 2hν -hν = hν \)

⇒ mV² = 2 hν 

\(⇒ V = \sqrt{\frac{2h\nu}{m}}\)

• Hence, option 2 is the answer