Question

Calculate the de Broglie wavelength in picometers (1 picometer =10^−12 meters) of an electron that has kinetic energy of 10 keV. The mass of an electron is 9.11 x 10^−31 kg

Answer 1

@chmvijay

Answer 2

@ganeshie8

Answer 3

@myininaya

Answer 4

@Compassionate

Answer 5

@nincompoop

Answer 6

@skullpatrol

Answer 7

@skullpatrol

Answer 8

@oOKawaiiOo

Answer 9

The wavelength is given by
\[ \lambda = \frac{h}{p}\]
Using the kinetic energy with the relativistic energy equations you get
\[K=E-mc^2\]
where
\[E = \gamma mc^2\]
and
\[\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\]
giving
\[K=\gamma mc^2 - mc^2
\\ \quad =mc^2(\gamma -1)
\\ \
\\ \quad K= mc^2 \left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right)\]
which lets you solve for the velocity, v of the electron since you're given kinetic energy and rest mass of the electron.

If you keep the velocity as a fraction of c until the final calculation, your life will be much easier!

Then, again, our original equation for the wavelength is
\[\lambda = \frac{h}{p}\]
where relativistic momentum is
\[p=\gamma mv\]
leaving you with
\[\lambda = \frac{h}{\gamma mv}
\\ \
\\ \quad =\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}\]

Answer 10

A useful conversion for either your mass or your kinetic energy is
\[1eV = 1.602 \times 10^{-19}J\]
\[1eV/c^2 = 1.783 \times 10^{-36} kg\]

Answer 11

Well, that all is excessive because it's not actually going that fast.

It's probably sufficient to use
\[K=\frac{p^2}{2m}\]
to solve for p, then
\[ \lambda = \frac{h}{p}\]

Answer 12

Very nice. I was about to add a note that the velocity should be estimated first because it might be much less than c and no relativistic analysis would be needed.