Find typical energy of nuclei in solar core

Click For Summary
SUMMARY

The typical energy of nuclei in the solar core is calculated using the formula for thermal energy, yielding a value of approximately 3.1 x 10-17 J at a core temperature of 1.5 x 106 K. The Coulomb barrier energy, necessary for protons to overcome repulsion, is calculated as 2.3 x 10-13 J. This analysis demonstrates that the thermal energy at the solar core temperature is insufficient to overcome the Coulomb barrier, highlighting the necessity of quantum tunneling for fusion reactions to occur. The temperature required to break the Coulomb barrier is estimated to be around 1010 K.

PREREQUISITES
  • Understanding of thermal energy calculations
  • Familiarity with Coulomb's law
  • Knowledge of quantum tunneling principles
  • Basic concepts of nuclear fusion
NEXT STEPS
  • Research quantum tunneling in nuclear fusion processes
  • Study the implications of temperature on nuclear reactions
  • Explore advanced thermal energy equations in astrophysics
  • Learn about the role of the strong nuclear force in fusion
USEFUL FOR

Students studying astrophysics, physicists interested in nuclear fusion, and researchers exploring stellar processes.

ck99
Messages
56
Reaction score
0

Homework Statement



Calculate the typical energy of nuclei in solar core. Calculate thermal energy required for a proton to overcome the Coulomb barrier and get within range of strong nuclear force (10^-15m)

Homework Equations



None given

The Attempt at a Solution



I have the core temp as 1.5 x 10^6 K. I know that typical nuclei will have mass mp which is 1.67 x 10^-27kg.

I am sure there must be a simple equation to work this out, but I've never come across it before! I know that the point of the question is to illustrate that fusion reactions would not occur at this temp without quantum tunneling. (Eg typical thermal energy is NOT enough to overcome Coulomb barrier.)
 
Physics news on Phys.org
A cup of tea and a browse of wikipedia has led me to the following attempt:

Typical energy = 3/2 kbT = 3.1 x 10-17J

Coulomb repulsion barrier energy = ke2/r = 2.3 x 10-13J

Rearrange the first formula to find that the temperature required to break the Couloumb barrier is approx 10^10K.

Is that the correct approach? (Feel embarrassed about posting this now, I knew it would be simple!)
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
Replies
35
Views
6K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K