# Electron/positron and photon

1. Apr 3, 2014

### johann1301

1. The problem statement, all variables and given/known data

A gammaphoton with a frequency of 5,49*10^20 Hz passes by near a nucleus, and turns into a electron/positron pair. The electron has an kinetic energy of 1,20*10^-13 J. How much kinetic energi is in the posistron?(we can neglect the energyexchange with the nucleus)

3. The attempt at a solution

Ive tried to solve this using this formula but i get a negative answer:

hf=2E+Ekp+Eke => Ekp=hf-2E-Eke

Where Ekp is the kinetic energi of the positron, "hf" is the energy of the photon, "E" is the rest energy of the electron and "Eke" is the kinetic energy of the electron.

I get that Ekp=-1,64*10^-10 J. This isnt right?

2. Apr 3, 2014

### dauto

The formula is correct. You must've made a mistake in your calculation.

3. Apr 3, 2014

### lightgrav

your mistake is in the Units ... 1 Joule is 1 kg m2/s2 , so your electron mass must be in kg.

4. Apr 4, 2014

### johann1301

As far as i can understand, the energy in the photon is not enough to create a electron/positron pair?

hf<2E

And therefore the problem doesnt have an answer?

5. Apr 4, 2014

### Curious3141

You're wrong, the problem does have an answer, which I just found. So we can be sure the problem is correct.

Your equation is also correct, so the error must lie in your working.

Unless you show this working, in detail, noone can help you further.

6. Apr 4, 2014

### johann1301

Ekp=hf-2E-Eke

Ekp=6,63*(10^-34)Js*5,49*(10^20)Hz - 2*9,1094*(10^-31)kg*(3,00*(10^8)m/s)^2 - 1,20*(10^-13)J

Ekp=8,00178*10^-14J

I just kept entering the wrong exponent in C^2.

Thanks:)

7. Apr 4, 2014

### Curious3141

What I got too.