Question

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

Answer 1

Correct Answer - Option 4 : \(2\sqrt {hv/m}\)

CONCEPT:

• Photoelectric Effect: Certain metals eject free electrons when a ray of light strikes them. This phenomenon is known as the Photoelectric effect.
• Einstein equation for the photoelectric effect is given as 

K = hν - ϕ 

ν is the frequency of light striking metal surface, h is Planck's constant, ϕ is work function.

• Work function (ϕ): The minimum energy required to emit electrons out of metal is known as the work function.
• Cutoff Frequency: The minimum frequency required by a ray of light to emit electrons is known as the Cutoff Frequency.

ϕ = hν₀

where ν₀ is the cutoff frequency for a light ray to emit electron out of given metal.

Kinetic Energy: The energy possed by a body in motion is known as Kinetic Energy.

\(K = \frac{1}{2} mv^2 \)

m is mass of the body, v is the velocity of the body

CALCULATION:

Given

Cuttoff frequency =  ν 

Frequency of light ray striking = 2ν 

The energy of each photon falling = 2hν 

the Maximum kinetic energy K obtained from the metal having cutoff frequency is v.

K = 2hν - hν (By Einestein Equation)

⇒ K = hν 

⇒ \(K = \frac{1}{2} mv^2 = h\nu\)

⇒ \(v^2 = \frac{ 2h\nu}{m}\)

⇒ \(v =\sqrt{\frac{ 2h\nu}{m}} \)