When two monochromatic lights of frequency, v and are incident on a photoelectric metal, their stopping potential becomes and Vs respectively. The threshold frequency for this metal is:

  1. 2v
  2. 3v

Answer (Detailed Solution Below)

Option 1 :
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Detailed Solution

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CONCEPT: 

  • When the photons fall on a metal surface then some electrons get ejected from the metal surface.This phenomenon is called the photoelectric effect.
  • The minimum energy needed to remove electrons from the metal surface is called work function (φ) of that metal.
  • The maximum energy of ejected electrons from the metal surface after ejection is called as maximum kinetic energy (KEmax).
  • Einstein’s equation of photoelectric equation:

Where ν = frequency of incident energy of photons, νo = threshold frequency, and KE = the maximum kinetic energy of electrons.

Here Wo = hνo 

If Vo is the stopping potential, e is the electronic charge, then

KEmax = eVo

⇒ eVo = (hν - hνo)

⇒ hν = hν+ eVo 

Calculation:

Frequency of first monochromatic light, ν1 = ν 

Stopping the potential of the first monochromatic light, V1 =  

We know, 

Then,  hν = W + eV1      

⇒        ---- (1)

Frequency of second monochromatic light, ν2 = ν/2

Stopping potential  second monochromatic light, V2 = Vs 

Then, 

         ---- (2)

Equate the value of W,

⇒ Vs = - h ν/e 

Put this equation 1,

W = 1.5 hν 

0 = 1.5 hν 

 

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