Estimate of Temperature Inside the Nucleus

A nucleus is about  
  m across. Any particle confined to the nucleus must have a wavelength of this order of magnitude. If we consider an  
- particle confined to the nucleus to have a wavelength of  
  m, then we can find the momentum using Dirac's wave particle duality equation  
\[p=\frac{h}{\lambda}=\frac{6.626 \times 10^{-34}}{10^{-15}}=6.626 \times 10^{-19}\]
The mass of an alpha particle is  
\[6.644 \times 10^{-27}\]
  kg, so  
\[p=mv \rightarrow v=\frac{p}{m}=\frac{6,626 \times 10^{-19}}{6.644 \times 10^{-27}} = 9.973 \times 10^{7}\]
This is fast - about one third the speed of light. The kinetic energy of the alpha particle is about  
\[\frac{1}{2}mv^2=\frac{1}{2} \times 6.644 \times 10^{-27} \times (9.973 \times 10^7)^2=3.3 \times 10^{-11}\]
  J or  
\[\frac{3.3 \times 10^{-11}}{1.6 \times 10^{-19}}=2.06 \times 10^8\]
Treating the alpja particle as a gas particle inside the nucleus - a big assumption, we can write  
\[kinetic \: energy=\frac{3}{2} kT \rightarrow T=\frac{3.3 \times 10^{-11}}{3/2 \times 1.38 \times 10^{-23}} = 1.59 \times 10^{12}\]
  degrees Kelvin.

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