Answer
a, 1.0 rad.
b. $1.0\times10^{10}\;protons$.
Work Step by Step
a. The RBE for protons is 2, according to Table 31-1, but for this problem we are told to assume that the RBE $\approx 1$.
The effective dose (in rem) is the absorbed dose (in rad) multiplied by RBE (equation 31-10a).
Here, the effective dose is 1.0 rem, so the absorbed dose is 1.0 rad or 0.010 Gy.
b. A Gy is 1 J/kg.
$$(0.010\;Gy)\frac{1J/kg}{Gy}(0.20kg)\frac{1\;eV}{1.60\times10^{-19}J}\frac{1p}{1.2\times10^6 eV}$$
$$=1.0\times10^{10}\;protons$$
