1.

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» 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 bya ray of light to emit electrons is known as theCutoff Frequency.

ϕ = hν0

whereν0 is the cutoff frequencyfor 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 havingcutoff frequency is v.

K = 2hν - hν (By Einestein Equation)

⇒ K = hν

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

\(v^2 = \frac{ 2h\nu}{m}\)

\(v =\sqrt{\frac{ 2h\nu}{m}} \)



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