Answer
$\beta=0.707$
Work Step by Step
Relativistic momentum is defined as $$p=\gamma mv$$ where $\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$ This means that if $p=mc$, then $$mc=\gamma mv$$ Cancelling out $m$ gives $$c=\gamma v$$ Using $\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$, $$c=\frac{v}{\sqrt{1-(\frac{v}{c})^2}}$$ Squaring each side gives $$c^2=\frac{v^2}{1-(\frac{v}{c})^2}$$ Cross multiplying yields $$c^2-v^2=v^2$$ Solving for $v$ gives $v=\frac{c}{\sqrt{2}}=.707c$
Since $v=\beta c$, $\beta=0.707$.
