## Estimate of Temperature Inside the Nucleus

A nucleus is about
$0^{-15}$
m across. Any particle confined to the nucleus must have a wavelength of this order of magnitude. If we consider an
$\alpha$
- particle confined to the nucleus to have a wavelength of
$10^{-15}$
m, then we can find the momentum using Dirac's wave particle duality equation
$p=\frac{h}{\lambda}=\frac{6.626 \times 10^{-34}}{10^{-15}}=6.626 \times 10^{-19}$
m.
The mass of an alpha particle is
$6.644 \times 10^{-27}$
kg, so
$p=mv \rightarrow v=\frac{p}{m}=\frac{6,626 \times 10^{-19}}{6.644 \times 10^{-27}} = 9.973 \times 10^{7}$
m/s.
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
$\frac{3.3 \times 10^{-11}}{1.6 \times 10^{-19}}=2.06 \times 10^8$
eV.
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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