Minimum wavelength when electrons strike phospor

1. Oct 5, 2009

songoku

1. The problem statement, all variables and given/known data
Assume electrons are accelerated through a potential difference of 20000 V inside a TV picture tube. What is the minimum wavelength that could be produced when the electrons strike the phosphor?

2. Relevant equations
W = qV

3. The attempt at a solution
I use the equation above to find the speed, then plug it in de Broglie formula to find the wavelength but there is no such answer in the multiple choices....Maybe we need to know the work function of phospor?

Thanks

2. Oct 5, 2009

kuruman

To get help, please show your work and the possible answers. Otherwise, we can't tell what you did wrong, if anything.

3. Oct 6, 2009

songoku

Hi kuruman

a. 0.62 angstrom
b. 1.15 angstrom
c. 10.5 angstrom
d. 100.9 angstrom

My work :

$$W=qV$$

$$\frac{1}{2}mv^2=qV$$

$$v=8386276.94~ms^-1$$

Then

$$\lambda = \frac{h}{mv}$$

$$=8.68\times 10^-12$$

$$=0.087~\text{angstrom}$$

4. Oct 6, 2009

kuruman

What you calculated here is the wavelength of a 20 keV electron, not the photon that it produces. What is the relation between energy and wavelength for a photon? You can use the de Broglie expression, but momentum is not mv for a photon. What is it?

5. Oct 7, 2009

songoku

Hi kuruman

Momentum for a photon = E/c , where E = hf

Then, the equation will be :

$$p=\frac{E}{c}=\frac{hf}{c}=\frac{h}{\lambda}$$

$$\lambda = \frac{h}{p}$$

I don't know how to find the momentum.....

Thanks

6. Oct 7, 2009

kuruman

You don't really need the momentum. You know that the energy of the photon is 20 keV and you are looking for its wavelength. You already have the expression

E/c = h/λ

so go for it.

7. Oct 7, 2009

songoku

Hi kuruman

Oh now I get it. Thanks a lot !