Question

Through how many volts of potential difference must an electron be accelerated to achieve a wavelength of 0.57 nm ?

Answer 1

We must use two formulas of energy first how energy is related to wavelength:

$E=\frac{hc}{\lambda}$

Such that $\lambda=Wavelength$, c is the speed of light in a vacuum and h is Plank's constant.

And the second equation is how energy relates to voltage:

$E=qV$

Such that q is the charge of the particle (in this case the electron) and V is voltage.  By substituting the second equation into the first we have:

$E=\frac{hc}{\lambda}\\ \\qV=\frac{hc}{\lambda} \\\\V=\frac{hc}{\lambda q}$

We know that:

$c=2.99 \times 10^8 m/s\\\\q=e=1.6 \times 10^{-19} C\\\\h= 6.67 \times 10^{-34} m^2kg/s\\\\\lambda=0.57 \times 10^{-9}m$

And so:

$V=\frac{hc}{\lambda q}=\frac{(6.67 \times 10^{-34})(2.99 \times 10^8)}{(0.57 \times 10^{-9}) (1.60 \times 10^{-19})}$

$V=2186.77 Volts$