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`
` <br>`
and `lambda_0 = 4lambda` .
The threshold wavelength for photoelectric emission is choice 2, `4lambda` .
No comments:
Post a Comment