Answer
6%
Work Step by Step
-Need to use these equations:
t($_{1/2}) = $$\frac{ln(2)}{k}$ and ln(N$_{t}$) = ln(N$_{0}$) - kt.
-Use the half life formula to find k.
t($_{1/2}) = $$\frac{ln(2)}{k}$ = 12.3yr$^{-1}$
k = 0.0564 yr$^{-1}$
-Solve for Nt by plugging in the found and given values.
ln(N$_{t}$) = ln(${3.024g}$) - [(0.0564yr$^{-1}$)(50yr)]
ln(N$_{t}$) = -1.71
e$^{ln(N_{t})}$ = e$^{-2.71}$
N$_{t}$ = 0.181g
-Divide by original mass and multiply by 100%
$\frac{0.181g}{3.024g}$ = 0.0598
3.024g is the mass of H multiplied by 3.
0.0598 $\times$ 100% = 5.98%
Therefore $\approx$ 6% of the tritium remains meaning that the dial will be dimmed by 94%
