I want to understand a concept better. I did a homework problem where I solved it all the way, then checked my answer with a solution set. My answer was different, so I followed the solution set from the beginning. At the beginning it says: "$\frac{\lambda}{\lambda_c}=0.2$ is relativistic for an electron."

The problem:

What must be the kinetic energy of an electron if the ratio of its de Broglie wavelength to its Compton wavelength is $0.2$?

What does this statement mean? In class, my professor always mentions that "[this example] is a non-relativistic situation", but I'm not sure what he meant by it.

My research: I googled some stuff and what I gathered is that there is a threshold of some sort when one must begin to consider the problem in a relativistic sense as opposed to a non-relativistic sense. This threshold is $1\%$, although I'm also not sure what this means either........


1 Answer 1


There is an important factor $\gamma = \dfrac {1}{\sqrt{1- \frac {v^2}{c^2}}}$ which crops up in relativity where $v$ is the speed of a particle and $c$ is the speed of light.

When the speed of a particle is very small compared with the speed of light this factor is very near to one.

However as the speed of a particle increases the factor $\gamma$ also increases.

When $\gamma$ is greater that $1.01$ (here is your $1\%$)then the particle is said to be relativistic and this occurs when $v\ge 0.14 c$.
Others may say that a particle is relativistic when $v\ge 0.1 c$.

It is just a rough and ready approximation to show when $\gamma$ starts to affect some of the equations of Classical Physics.


$\lambda_C = \dfrac {h}{mc}$ and $\lambda_{DB} = \dfrac {h}{\gamma mv}$
Note the presence of the factor $\gamma$ because you need to consider relativistic effects.

$\Rightarrow \dfrac{\lambda_C}{\lambda_{DB}} = \dfrac{\gamma v}{c}$ with $\gamma = \dfrac {1}{\sqrt{1- \frac {v^2}{c^2}}}$

Rearrange to get $\dfrac v c$ in terms of $\dfrac{\lambda_C}{\lambda_{DB}}$

This should give a value of $\dfrac v c = 0.98$ and $\gamma = 5.1$ which shows that using the equations of Classical Physics will not produce the correct answer.

Finally the kinetic energy is $(\gamma -1) m c^2$

  • $\begingroup$ When I solve for $v$ it comes out to $5c$, so I don't think this is the way to get $\gamma$. The solution set uses $\frac{u/c}{\sqrt{1-(u/c)^2}}=\frac{\lambda_c}{\lambda}$ and then proceeds to solve for $u/c$. From there they skip a step and simply say that $\gamma=5.10$. I don't understand... $\endgroup$ Commented Feb 14, 2016 at 17:36
  • $\begingroup$ @whatwhatwhat I have added a little more to my answer which I hope will help you. $\endgroup$
    – Farcher
    Commented Feb 14, 2016 at 22:55
  • $\begingroup$ So we always need to calculate $\gamma$ or the velocity of the particle in order to determine if it is relativistic or not? The solution set simply determined it by the ratio of wavelengths. I mean, I know it's a solution set but still...is there a way to know from the ratio of these wavelengths? $\endgroup$ Commented Feb 16, 2016 at 0:33

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