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I was studying experiment on photoelectric effect. It was about effect of light intensity, frequency, potential etc. on photocurrent and KE of the ejected electrons (referring to the below setup). The book never considered energy loss because of accelerating electrons by radiation in the glass tube. (As accelerating charges produce em waves and therefore there is an energy loss)

It this for convenience or is there something involved i am not getting it or is it that energy loss is negligible so it is ignored?

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In an absolute vacuum there won't be any energy loss of the photoelectrons in the glass tube. They are uniformly accelerated from the cathode to the anode. Energy loss only happens if you have a sufficient gas density in the tube. You will notice this then as a glow appearing in the form of striations caused by collisional excitation of the gas molecules.

The radiative energy loss due to the acceleration itself can be completely neglected in case of a linear acceleration. It is only a fraction

$$q=\frac{2e^2}{3m^2c^3v}\cdot\frac{dE}{dx}$$

of the energy gained by the acceleration, where $e$ is the elementary charge, $m$ the electron mass, $v$ its velocity, $c$ the speed of light and $dE/dx$ the potential energy gradient of the field (all in $cgs$-units) (see Eq.50 in this reference)

If you assume a field of the order of $10V/cm$ this results in values for $q$ of the order of $10^{-15}$, which will not be measurable.

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  • $\begingroup$ shouldn't accelerating electrons produce em radiations and loose energy?? $\endgroup$
    – Zitscx ø
    Jan 24, 2022 at 8:31
  • $\begingroup$ sry forgot to put energy loss by radiation in the question. I have edited it. $\endgroup$
    – Zitscx ø
    Jan 24, 2022 at 9:21
  • $\begingroup$ @Zitscxø Radiative losses are completely negligible in case of linear acceleration. See my edited answer. $\endgroup$
    – Thomas
    Jan 24, 2022 at 19:44
  • $\begingroup$ Thanks for answering. Cleared my doubt.. $\endgroup$
    – Zitscx ø
    Jan 25, 2022 at 5:16

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