GRE Physics GR 1777 Problem 036 Solution

You should know

Energy-momentum Relation

$$ \large E^2 = (pc)^2 + (mc^2)^2 $$

where

$E$: total energy (relativistic energy)

$p$: momentum of magnitude

$m$: rest mass (invariant mass)

$c$: speed of light

Solution

Total energy $E$ = 5.0 GeV

Momentum of magnitude $p$ = 4.9 GeV/c

Then,

$$ \large \begin{split} (m c^2)^2 & = E^2 - (pc)^2 \\ \\ & = (5.0 \text{ GeV})^2 - (4.9 \text{ GeV})^2 \\ \\ & = 25 - 24.01 \text{ (GeV)}^2 \\ \\ & = 0.99 \text{ (GeV)}^2 \end{split} $$

Therefore, the rest mass of the particle is approximately

$$ \large m \simeq 1.0 \text{ GeV/c}^2 $$

Answer

(D) 1.0 GeV/c$^2$

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