Question

What will be energy of proton if it is moving with 97% speed of light?

Answer 1

Put that $\alpha=0.97=v/c.$ Then the energy of such a photon will be

$$E=\sqrt{p^2c^2+(m_0c^2)^2}=\sqrt{\frac{m_0^2\alpha^2c^4}{1-\alpha^2}+(m_0c^2)^2}=\\\space\\ =\frac{m_0c^2}{\sqrt{1-\alpha^2}}=\frac{1.673\cdot10^{-27}(3\cdot10^8)^2}{\sqrt{1-0.97^2}}=1.671\cdot10^{-17}\text{ J},$$

or 104.3 eV.