1
$\begingroup$

Show that when a photon of energy E is scattered from a free electron at rest, the max kinetic energy of the recoiling electron is given by

$$K_\text{max}=\frac{E^{2}}{E+\frac{mc^{2}}{2}}$$

Using $$\Delta \lambda = \frac{h}{mc}\left ( 1-\cos\phi \right )$$, the largest change in energy transfer from the photo to the electron, by conservation, would result in a larger change in wavelength.

Change in wavelength largest when $$\phi=\pi$$

Finally, the expression becomes

$$\Delta \lambda=\lambda_{f}-\lambda_{i}=\frac{2h}{m_{e}c}$$

Stuck here.

Can someone give me a little help?

Edit:Tried substituting delta lambda into $$\Delta E=\frac{hc}{\Delta \lambda}$$ to figure out the change in energy which would be the energy gained by the electron after the collision but arrived at nothing similar to the required expression.

Edit:

I see what's the problem-technical algebraic error.

happy for this topic to be deleted .

$\endgroup$
6
  • 1
    $\begingroup$ note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Dec 10 '15 at 11:39
  • $\begingroup$ @JohnRennie I am aware. But I am confused as to how my question does not fit the guidelines. I don't think this question is a high-level question either. It's a review of some topics that I did in my first year undergrad. $\endgroup$ – Physkid Dec 10 '15 at 11:43
  • $\begingroup$ Change in wavelength is maximum at $\phi=\pi$ and not at $\phi=\pi/2$ $\endgroup$ – Muhsin Ibn Al Azeez Dec 10 '15 at 11:43
  • $\begingroup$ @MuhsinIbnAlAzeez It's a typo. But this error does not change any of the calculation above. $\endgroup$ – Physkid Dec 10 '15 at 11:44
  • $\begingroup$ You could try using linear momentum conservation for the electron and photon using the velocity components on both axes. $\endgroup$ – Tamoghna Chowdhury Dec 10 '15 at 13:37
3
$\begingroup$

First of all, as this is a homework type quertion, I don't post full answer.

You can't write $\Delta{E}=\frac{hc}{\Delta\lambda}$ because $\Delta\left( \frac{1}{\lambda}\right)=\frac{-1}{\lambda^2}\Delta\lambda$

Now, you can simply write, from compton's equation, $$\lambda'-\lambda=\frac{2h}{mc}\\ \implies \frac{1}{E'}-\frac{1}{E}=\frac{2}{mc^2}$$
This get when divide throuht by $hc$. Here, $E'$ and $\lambda'$ are energy and wavelength of defracted wave length. You can use conservation of energy to replace $E'$ in terms of $E$ and kinetic energy of electron. Put that into above equation and rearranging, you get the required relation.

[edit]: as per comments, I add a little more. You cannot write $\Delta{E}=\frac{hc}{\Delta\lambda}$ because , $E=\frac{hc}{\lambda}$ and so $\Delta{E}=\Delta\left(\frac{hc}{\lambda}\right)=\frac{-hc}{\lambda^2}\Delta\lambda$ .
And also the expression $\Delta{E}=\frac{hc}{\Delta\lambda}$ does not make any physical sense. Because this means that larger the difference between wavelength of incident and scattered photon, smaller the energy gained by the electron. And also, as the wavelength increases, energy decreases. So, there must be a negative sign.

Hope you understand it.

$\endgroup$
6
  • $\begingroup$ Spot-on with your second line ("You can't write...")! $\endgroup$ – Floris Dec 10 '15 at 13:43
  • $\begingroup$ I chuckled at the second line. That was a real embarrassment. $\endgroup$ – Physkid Dec 10 '15 at 14:16
  • $\begingroup$ I have used the conservation law : $$E-K=E'-K'$$ Where E is the energy of photon before collision and the prime denotes after. Where K is the energy of electron before collision (k=0) and the primed K is the electron's energy after collision. expressed the above in terms of E' and subbed into the prevailing equation "1/E'-......". But it turned out futile. Is this at this point a matter of algebraic manipulation? Prob wouldn't want to waste time on manipulations. $\endgroup$ – Physkid Dec 10 '15 at 15:01
  • $\begingroup$ @Physkid how you get that relation? total energy before collision is $E+K$ and after collision is $E'+K'$ thus you have to write $E+K=E'+K'$ with $E=0$ use this and try again $\endgroup$ – Muhsin Ibn Al Azeez Dec 11 '15 at 12:49
  • $\begingroup$ @MuhsinIbnAlAzeez Typos again! The 'minus' was meant to be a 'Plus' but otherwise I had the same thought and equation as you did. Why would E' =0 as you put it? Certainly, K=0 because initially the electron is at rest and from classical mechanics K=0 IFF v=0. But E=0? How can the initial energy of the photon be zero when the photo is moving with speed of light c? $\endgroup$ – Physkid Dec 11 '15 at 13:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.