Momentum in a photon -> cathode collision

[ChessEnthusiast]

Homework Statement

A photon with the length of lambda hits a cathode perpendicularly to its surface. As a result, an electron leaves its surface perpendicularly to the direction of the photon. How much momentum was transferred from the photon to the cathode? The work function of the cathode is W.

2. The attempt at a solution
1. I calculate the speed of the electron:
$$v = \sqrt{ \frac{2hc}{\lambda m_e} - \frac{W}{m} }$$

2. The momentum of the photon:
$$p_{photon} = \frac{h}{\lambda}$$

3. The momentum of the electron:
$$p_{electron} = m \sqrt{ \frac{2hc}{\lambda m_e} - \frac{W}{m} }$$

Now, I am stuck and don't know what to do because of this 90 degree angle.

 

[kuruman]

Conserve momentum in two dimensions, perpendicular and parallel to the cathode surface.

Note: your expression for v is incorrect.

 

[ChessEnthusiast]

kuruman,
Thank you, there is a 2 in front of W in the velocity expression.

Could you give me some hints how to conserve this momentum? The only thing that comes to my mind is
$$p_{cathode} + p_{electron} = - p_{photon}$$

 

[kuruman]

That's it, but you write it as a vector equation. Call the perpendicular direction y and the parallel direction x. What are the x and y components oh the three momentum vectors?

On edit: Lose the negative sign. The total momentum vector before is equal to the total momentum vector after.

 

[ChessEnthusiast]

in x direction:
$$p_{x cathode} = p_{electron}$$

in y direction:
$$p_{y cathode} = p_{photon}$$

$$p_{cathode} = \sqrt{p_{x cathode}^2 + p_{y cathode}^2}$$

 

[kuruman]

Something like that. It's not clear if the problem is asking for the vector or its magnitude which is what you found.