For photoelectric emission from certain metal the cut off frequency is

1.	For photoelectric emission from certain metal the cut off frequency is v. If radiation of frequency 2v impinges onthe 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» Correct Answer - Option 2 : \(\sqrt{\dfrac{2h\nu}{m}}\) CONCEPT: Thework function istheminimum amount of energy required to releasetheelectronfrom a photoemissive surface and is given by ⇒ ϕ = hν0 Where h = Plancks constant,ν0= threshold frequency TheEnsitens photoelectricequationgivesthekinetic energy of a photoelectron, thekinetic energy ofaphotoelectron is the difference in energy betweentheincident photon and work functionof the material and is given by ⇒Kinetic energy(KE) = hν -ϕ Whereν = 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 frequencyν ) If the frequency is doubled then the above equation can be written as \(⇒ \frac{1}{2}mV^{2} = 2hν -hν_{0}\) Assumeν =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

For photoelectric emission from certain metal the cut off frequency is v. If radiation of frequency 2v impinges onthe 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» Correct Answer - Option 2 : \(\sqrt{\dfrac{2h\nu}{m}}\)

CONCEPT:

Thework function istheminimum amount of energy required to releasetheelectronfrom a photoemissive surface and is given by

⇒ ϕ = hν0

Where h = Plancks constant,ν0= threshold frequency

TheEnsitens photoelectricequationgivesthekinetic energy of a photoelectron, thekinetic energy ofaphotoelectron is the difference in energy betweentheincident photon and work functionof the material and is given by

⇒Kinetic energy(KE) = hν -ϕ

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

EXPLANATION:

⇒KE = hν -ϕ

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

\(⇒ \frac{1}{2}mV^{2} = 2hν -hν_{0}\)

Assumeν =ν0, the above equation can be written as

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

⇒ mV2= 2 hν

\(⇒ V = \sqrt{\frac{2h\nu}{m}}\)