Calculating Kinetic Energy of a Positron from Gamma Photon Interaction

  • Thread starter Thread starter johann1301
  • Start date Start date
  • Tags Tags
    Photon
Click For Summary

Homework Help Overview

The discussion revolves around calculating the kinetic energy of a positron resulting from the interaction of a gamma photon with a nucleus, specifically when the photon frequency is given. The original poster attempts to determine the kinetic energy of the positron after accounting for the kinetic energy of the electron produced in the pair creation process.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the validity of the formula used for calculating the positron's kinetic energy and question the calculations leading to a negative result. There are inquiries about the energy of the photon being sufficient for pair production.

Discussion Status

Some participants have provided guidance on potential calculation errors and emphasized the importance of unit consistency. There is an ongoing exploration of whether the energy from the photon is adequate for creating an electron-positron pair, with differing opinions on the problem's solvability.

Contextual Notes

Participants note that the energy of the photon may not meet the threshold required for pair production, leading to discussions about the implications of this on the problem's validity. There is also mention of the need for detailed working to assist in identifying errors in calculations.

johann1301
Messages
216
Reaction score
1

Homework Statement



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)

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 isn't right?
 
Physics news on Phys.org
The formula is correct. You must've made a mistake in your calculation.
 
your mistake is in the Units ... 1 Joule is 1 kg m2/s2 , so your electron mass must be in kg.
 
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 doesn't have an answer?
 
johann1301 said:
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 doesn't have an answer?

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, no one can help you further.
 
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:)
 
johann1301 said:
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:)

What I got too.

Here's a tip: Google calculator is your friend. Take a look at this: https://www.google.com.sg/search?q=...+-+2*mass+of+electron*c^2+-+1.2E-13J&safe=off

Just enter your expression as a query next time and Google works everything out, physical constants, units, etc. all inclusive!
 
Thanks!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K