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= (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= (h)/(\lambda)= (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= (p^2)/(2m)= ((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= (E)/(q)= (6.99 \cdot 10^(-18) J)/(1.6 \cdot 10^(-19) C)=43.7 V`