Question

Through what potential difference δv must electrons be accelerated (from rest) so that they will have the same wavelength as an x-ray of wavelength 0.185 nm ?

Answer 1

The De Broglie wavelength of the electron is equal to

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

where h is the Planck constant and p is the electron's momentum. Since this wavelength must be equal to that of the x-ray,

$\lambda=0.185 nm=0.185 \cdot 10^{-9} m$

we can re-arrange the previous equation to find the momentum of the electron:

$p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34}Js}{0.185 \cdot 10^{-9} m}=3.57 \cdot 10^{-24} kg m s^{-1}$

The kinetic energy of the electron is equal to the square of the momentum divided by twice its mass:

$E= \frac{p^2}{2m}= \frac{(3.57 \cdot 10^{-24}kgms^{-1})^2}{2 (9.1 \cdot 10^{-31} kg)} =6.99 \cdot 10^{-18} J$

When the electron is accelerated by a potential difference $\Delta V$, the energy it gains is

$E=q \Delta V$

where q is the electron charge. Re-arranging the formula, we find the potential difference:

$\Delta V= \frac{E}{q}= \frac{6.99 \cdot 10^{-18} J}{1.6 \cdot 10^{-19} C}=43.7 V$