Answer
$\text { From } \gamma=1+K / m c^{2}(\text { see } \mathrm{Eq} .37-52) \text { and }$ $$v=\beta c=c \sqrt{1-\gamma^{-2}}(\text { see Eq. } 37-8),$$
we get
$$v=c \sqrt{1-\left|1+\frac{K}{m c^{2}}\right|^{-2}} $$
Therefore, for the $\Sigma^{* 0}$ particle,
$$v=\left(2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\right) \sqrt{1-\left(1+\frac{1000 \mathrm{MeV}}{1385 \mathrm{MeV}}\right)^{-2}}=2.4406 \times 10^{8} \mathrm{m} / \mathrm{s} $$
For $\Sigma^{0} $
$$v^{\prime}=\left(2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\right) \sqrt{1-\left(1+\frac{1000 \mathrm{MeV}}{1192.5 \mathrm{MeV}}\right)^{-2}}=2.5157 \times 10^{8} \mathrm{m} / \mathrm{s} $$
Thus $\Sigma^{0}$ moves faster than $\Sigma^{* 0}$.
Work Step by Step
$\text { From } \gamma=1+K / m c^{2}(\text { see } \mathrm{Eq} .37-52) \text { and }$ $$v=\beta c=c \sqrt{1-\gamma^{-2}}(\text { see Eq. } 37-8),$$
we get
$$v=c \sqrt{1-\left|1+\frac{K}{m c^{2}}\right|^{-2}} $$
Therefore, for the $\Sigma^{* 0}$ particle,
$$v=\left(2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\right) \sqrt{1-\left(1+\frac{1000 \mathrm{MeV}}{1385 \mathrm{MeV}}\right)^{-2}}=2.4406 \times 10^{8} \mathrm{m} / \mathrm{s} $$
For $\Sigma^{0} $
$$v^{\prime}=\left(2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\right) \sqrt{1-\left(1+\frac{1000 \mathrm{MeV}}{1192.5 \mathrm{MeV}}\right)^{-2}}=2.5157 \times 10^{8} \mathrm{m} / \mathrm{s} $$
Thus $\Sigma^{0}$ moves faster than $\Sigma^{* 0}$.
