Answer
$T_{1/2} = 1.12 \times 10^{11} \space year$
Work Step by Step
The equation of number of nuclei, $N $,
$N = \frac{R}{\lambda} = \frac{R}{ln2/T_{1/2}} = \frac{R (T_{1/2} )}{ln 2}$
Rearrange the equation to solve for half life, $T_{1/2}$, so the equation becomes
$T_{1/2} = \frac{N (ln2)}{R} $
To know the half life, we need to solve $N$ first. $0.15%$ of $1.00g$ is the mass of $^{147} Sm$. So, we use the equation for mass related to find number of nuclei, $N$.
$N = \frac{(0.150)(1.00g)}{147 u \times 1.661 \times 10^{–24} g/u}$
$N = 6.143 \times 10^{20}$
Now solve for $T_{1/2}$,
$T_{1/2} = \frac{(6.143 \times 10^{20}) (0.693)}{120 s^{-1}} $
$T_{1/2} = 3.55 \times 10^{18} s$ .
Change to year
$T_{1/2} = 1.12 \times 10^{11} \space year$
