When a piece of metal is illuminated by a monochromatic light of wavelength lambda, then stopping potential is 3Vs. When the same surface is illuminated by light of wavelength 2 lambda, then stopping potential becomes Vs. The value of threshold wavelength for photoelectric emission will be 1)8 lamda 2)4 lamda 3)6 lamda 4)4/3 lamda

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The stopping potential refers to the potential difference required to stop an electron emitted from a metal after it is illuminated by the light. This potential difference is equal to the kinetic energy with which electron leaves the metal:

`K = hf - phi` .

This is called photoelectric equation ...

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The stopping potential refers to the potential difference required to stop an electron emitted from a metal after it is illuminated by the light. This potential difference is equal to the kinetic energy with which electron leaves the metal:

`K = hf - phi` .

This is called photoelectric equation. Here, h is the Plank's constant, f is the frequency of the incident light and `phi` is the work function of the metal. The work function is related to the threshold frequency of the incident light: `phi = hf_0` , where `f_0` is the minimum frequency for which any electrons will be emitted.

The frequency of light can be expressed through its wavelength `lambda` as

`f = c/lambda` , where c is the speed of light.

In the given problem, the stopping potential for the light with wavelength `lambda`

is 3V. The photoelectric equation becomes

`3V =(hc)/lambda - phi`

The stopping potential for the light with wavelength `2lambda ` is V:

`V = (hc)/(2lambda) - phi`

These two equations can be solved together for the work function. Multiplying the second equation by -2 and adding it to the first equation results in

`V = phi`

The work function equals `phi = (hc)/lambda_0=V` , where `lambda_0` is the threshold wavelength. Combining this with second of the photoelectric equations, we get

`(hc)/lambda_0 = (hc)/(2lambda) - (hc)/lambda_0`

From here,

`(2hc)/lambda_0 = (hc)/(2lambda)` . Taking reciprocal in both sides results in

 

`lambda_0/2 = 2lambda`

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and `lambda_0 = 4lambda` .

The threshold wavelength for photoelectric emission is choice 2, `4lambda` .

 

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