# [Physics] Is the kinetic energy of an electron always $1.6 \cdot 10^{-19}~\text{J}$

energyhomework-and-exercisesmomentumopticswave-particle-duality

I was doing the following problem.

For $d$. They bizarrely added an extra step. They calculated the momentum of the electron to find the De Broglie wavelength using the kinetic energy relationship.

However, they assume the kinetic energy of the particle is $1.6 \cdot10^{-19}~\text{J}$. Is this always true? What do they mean by this? It doesn't seem to give me the same momentum as $mv=p$.

To answer the question in your title - no, the kinetic energy of an electron is a function of its velocity, same as for any other particle. The charge of an electron is always $1.6\cdot 10^{-19}C$, and so the electron will pick up $1.6\cdot 10^{-19}J$ of energy for every Volt of accelerating potential.

The key to answering the question (part d in the problem set) is the calculation of the wavelength of the electron.

The De Broglie wavelength is given by

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

Now the velocity is much less than the speed of light, so we don't need to consider relativistic effects. We know the rest mass of the electron, and we find

$$\lambda = \frac{6.6\cdot 10^{-34}}{9.1\cdot 10^{-31}\cdot 1.5\cdot 10^6}=0.5 nm$$

The other equation used ($p=\sqrt{2mK}$) appears to be using information not given in the question: namely that the kinetic energy of the electron is 5 eV, which is what you get when an electron is accelerated through 5 Volt. That has no visible bearing on the calculation of the wavelength, which can be derived directly from the velocity of the electron.

Incidentally, the momentum-from-energy calculation gives a completely different answer from the momentum-from-velocity equation, and indeed an electron with 5 eV of energy does not have a speed of 1.5E6 m/s... No wonder you were getting confused.