What does it mean to say that something is "relativistic for an electron"?

Question

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........

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.

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Later

$\lambda_C = \dfrac {h}{mc}$ 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 $(\gamma -1) m c^2$