Question

What is the de Broglie wavelength of an electron which is accelerated through a potential difference of 10 kV ?

Answer 1

The de Broglie wavelength `lambda=1.23` x `10^(-9)m`.

If a charge of 1 Coulomb is moved through a potential difference of 1 Volt then 1 Joule of work is done. This is how the Volt is defined.

So if a charge of `e` coulombs is moved through a potential difference of `V` volts then the work done is `eV` Joules.

In this case `e` is the charge on the electron which is equal to
`-1.6` x `10^(-19)C`.

The work done in accelerating an electron through this electric field will be equal to the kinetic energy of the electron `(1)/(2)mv^(2)`.

So we can write `eV=(1)/(2)mv^(2)`

From this we can work out the velocity of the electron which will then enable us to work out the de Broglie wavelength.

`v^(2)=(2eV)/(m)`

`v=sqrt((2eV)/(m)`

The mass of the electron `m=9.11` x `10^(-31)kg`

So `v=sqrt((2(1.6.(10^(-19)).10^(4)))/(9.11.(10^(-31))`

From which `v=5.9` x `10^(7)m.s^(-1)`

The de Broglie wavelength is given by:

`lambda=(h)/(mv)`

Where `h` is the Planck Constant which is equal to

`6.63` x `10^(-34) J.s`

So:

`lambda=(6.63(10^(-34)))/(9.11.(10^(-31))5.9.(10^(7))`

So we get :

`lambda`= 1.23 x `10^(-9)m`