# Electron potential/momentum problem

1. Sep 28, 2015

### orangeincup

1. The problem statement, all variables and given/known data
To produce x-ray radiation, electrons are accelerated in vacuum and aimed so as to collide with a target. Assume that all of the electron kinetic energy is converted into the energy of the x-ray photon on collision and that the wavelength of the resulting radiation is 1.24 angstroms.

2. Relevant equations
h^2/(2*(e)(m)(λ)^2)

3. The attempt at a solution
(6.63*10^-34)^2/(2(1.6*10^-19)(9.11*10^-31)(1.24*10^-9)^2)

=.979 V
/1000 for kV = .000979 kV. I entered this answer and it told me it was incorrect.

2. Sep 29, 2015

### andrevdh

That equation is unknown to me. Try starting out with the basic equation for the energy of the x-ray photon.

3. Sep 29, 2015

### ehild

What is the question?
You did not write any equation. What does the formula in 2 mean?

4. Sep 29, 2015

### orangeincup

The question is "What potential difference must the electron be accelerated? (answer in kV) "
Enter the electron momentum in kg*m/s

My equation I found using KE=1/2mv^2, λ = h / p, and I guess I messed something up

so I have the λ=1.26, the mass of an electron, the fact it is a vaccum, the charge of an electron, planks constant. What kind of formula should I be using?

h/ λ =p
(6.63*10^-34)/(1.24*10^-10) = 5.34*10^-24 (momentum?)

5. Sep 29, 2015

### ehild

You need a formula that contains the accelerating voltage.
The problem asks the momentum of the electron. h/ λ =p is right if you use the wavelength of the electron, but you substituted the wavelength of the photon, so you got the momentum of the photon.

6. Sep 30, 2015

### andrevdh

The kinetic energy of the electrons are used to produce the x-ray photons. So I suggest starting out with the energy of the x-ray photons. Do you know what the equation is for the energy of a photon (according to Planck and as used by Einstein to explain the photoelectric effect)?