- #1
Von Neumann
- 101
- 4
Problem:
Derive a formula expressing the de Broglie wavelength (in Å) of an electron in terms of the potential difference V (in volts) through which it is accelerated.
Solution (so far):
The textbook's answer is the following,
[itex]\lambda=12.27[V(\frac{eV}{2m_{0}c^{2}}+1)]^{-\frac{1}{2}}[/itex]
I'm having some trouble getting there. I started by noting that if an electron is accelerated from rest through a potential difference V, it gains a kinetic energy
[itex]\frac{1}{2}mv^{2}=eV[/itex]
So therefore,
[itex]v=\sqrt{\frac{2eV}{m}}[/itex]
Since [itex]\lambda=\frac{h}{mv}[/itex]
Therefore, [itex]\lambda=\frac{h}{\sqrt{2meV}}[/itex]
Also, [itex]\frac{h}{\sqrt{2me}}\approx 12.27 Å[/itex]
So I get, [itex]\lambda=\frac{12.27Å}{\sqrt{V}}[/itex]
I don't know where I went wrong. Any suggestions?
Derive a formula expressing the de Broglie wavelength (in Å) of an electron in terms of the potential difference V (in volts) through which it is accelerated.
Solution (so far):
The textbook's answer is the following,
[itex]\lambda=12.27[V(\frac{eV}{2m_{0}c^{2}}+1)]^{-\frac{1}{2}}[/itex]
I'm having some trouble getting there. I started by noting that if an electron is accelerated from rest through a potential difference V, it gains a kinetic energy
[itex]\frac{1}{2}mv^{2}=eV[/itex]
So therefore,
[itex]v=\sqrt{\frac{2eV}{m}}[/itex]
Since [itex]\lambda=\frac{h}{mv}[/itex]
Therefore, [itex]\lambda=\frac{h}{\sqrt{2meV}}[/itex]
Also, [itex]\frac{h}{\sqrt{2me}}\approx 12.27 Å[/itex]
So I get, [itex]\lambda=\frac{12.27Å}{\sqrt{V}}[/itex]
I don't know where I went wrong. Any suggestions?
Last edited: