Answer
$\beta=0.99989$
Work Step by Step
To find relativistic energy, use the equation $$E=\gamma mc^2$$ Solving for gamma and substituting a value of $m_p=1.6726\times 10^{-27}kg$ yields $$\gamma=\frac{E}{mc^2}=\frac{10.611\times 10^{-9}J}{(1.6726\times 10^{-27}kg)(3.00\times 10^8m/s)^2}=70.49$$ The relationship between gamma and beta is $$\gamma=\frac{1}{\sqrt{1-\beta^2}}$$ Taking the reciprocal of each side and squaring each side yields $$\frac{1}{\gamma^2}=1-\beta^2$$ This means that $$\beta=\sqrt{1-\frac{1}{\gamma^2}}$$ Substituting in a value of $\gamma=70.49$ gives a beta of $$\beta=\sqrt{1-\frac{1}{70.49^2}}=.99989$$
