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=(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=(hc)/(\lambda)\\ \\qV=(hc)/(\lambda) \\\\V=(hc)/(\lambda q)`

We know that:

`c=2.99 * 10^8 m/s\\\\q=e=1.6 * 10^(-19) C\\\\h= 6.67 * 10^(-34) m^2kg/s\\\\\lambda=0.57 * 10^(-9)m`

And so:

`V=(hc)/(\lambda q)=((6.67 * 10^(-34))(2.99 * 10^8))/((0.57 * 10^(-9)) (1.60 * 10^(-19)))`

`V=2186.77 Volts`