When the light of frequency 2v0 (where v0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5v0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is

  1. 1 ∶ 2
  2. 1 ∶ 4
  3. 2 ∶ 1
  4. 4 ∶ 1 

Answer (Detailed Solution Below)

Option 1 : 1 ∶ 2
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Detailed Solution

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Concept:

The phenomenon of ejection of electrons from metal surface when a light of frequency greater than the threshold frequency of it is incident on it is known as photoelectric effect.

Energy of incident photon is used to provide threshold energy to the electrons on the metal surface and the energy left is imparted to electron as its kinetic energy.

Equation connecting energy of incident photon and threshold energy and kinetic energy of ejected electron is 

\(E'=E_T+E_e\)

\(hv=hv_0+\frac{1}{2}{\rm mv}^2\)

Frequency of incident photon and threshold frequency of metal surface is v and vrespectively.

Calculation:

Given-

Threshold frequency = v0

Frequency of incident light = 2v0 

\(h\left( {2{\nu _0}} \right) = h{\nu _0} + \frac{1}{2}mv_1^2\)

\(h{\nu _0} = \frac{1}{2}mv_1^2\)      ----(i)           

\(4h{\nu_0} = \frac{1}{2}mv_2^2\)      ----(ii)           

Divide (i) by (ii),

\(\frac{1}{4} = \frac{{v_1^2}}{{v_2^2}}\)

\(\frac{{{v_1}}}{{{v_2}}} = \frac{1}{2}\)

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