Question

A gold nucleus (with a radius of 6.5 fm in its rest system, containing 197 protons and neutrons of rest mass 939 MeV/c2 each) is accelerated to a speed of 0.97 c. Calculate (a) the momentum of an individual nucleon (ignoring binding energy effects). (b) the total energy of the gold nucleus, including its binding energy of 7.9 MeV per nucleon. (c) the diameter of the gold nucleus in the direction of motion as seen by an observer at rest.

Answer 1

Answer:

$1.9982154567\times 10^{-18}\ kgm/s$

762474.685899 MeV

3.1603639031 fm

Explanation:

v = 0.97 c

c = Speed of light = $3\times 10^8\ m/s$

Relativistic momentum is given by

$p=\dfrac{m_0v}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow p=\dfrac{939\times 10^6\times 1.6\times 10^{-19}\times 0.97\times 3\times 10^8}{\sqrt{1-\dfrac{0.97^2c^2}{c^2}}}\\\Rightarrow p=\dfrac{(939\times 10^6\times 1.6\times 10^{-19})J/c^2\times 0.97c}{\sqrt{1-0.97^2}}\\\Rightarrow p=\dfrac{(939\times 10^6\times 1.6\times 10^{-19})\times 0.97}{c\sqrt{1-0.97^2}}\\\Rightarrow p=\dfrac{(939\times 10^6\times 1.6\times 10^{-19})\times 0.97}{3\times 10^8\times \sqrt{1-0.97^2}}\\\Rightarrow p=1.9982154567\times 10^{-18}\ kgm/s$

The momentum is $1.9982154567\times 10^{-18}\ kgm/s$

Energy in MeV

$E=\dfrac{m_0c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow E=\dfrac{939}{\sqrt{1-0.97^2}}\\\Rightarrow E=3862.52987766\ MeV$

Total energy is

$E'=3862.52987766+7.9=3870.42987766\ MeV$

The total energy is 3870.42987766 MeV

For all the nucleons

$E_t=197\times 3870.42987766=762474.685899\ MeV$

The energy is 762474.685899 MeV

Diameter = $2\times 6.5=13\ fm$

From length contraction

$D'=D\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow D'=13\sqrt{1-0.97^2}\\\Rightarrow D'=3.1603639031\ fm$

The diameter would be 3.1603639031 fm