^ ok, so considering the total energy:
E_{tot} = pc + (m_{0}c^{2})
ie, pc = E_{tot} - m_{0}c^{2}
So I should be able to use:
\lambda = \frac{hc}{pc} = \frac{hc}{E_{tot} - m_{0}c^{2}}
for the total energy expression?
This would give \lambda_{muon} = 4.21fm where m_{0,muon} =...