Answer
(a) $30\%$
(b) $23.6\%$
(c) $18.5\%$
Work Step by Step
(a) We know that
$\frac{N_1}{N_{\circ}}=e^{-(\frac{\ln 2}{T_{\frac{1}{2}}})t_1}$
We plug in the known values to obtain:
$\frac{N_1}{N_{\circ}}=e^{-(0.024062y)(50y)}=0.300=30\%$
(b) We know that
$\frac{N_2}{N_{\circ}}=e^{-(0.024068)(60y)}$
$\frac{N_2}{N_{\circ}}=0.236=23.6\%$
(c) We know that
$\frac{N_3}{N_{\circ}}=e^{-\frac{\lambda}{3}}$
We plug in the known values to obtain:
$\frac{N_3}{N_{\circ}}=e^{-(0.020468y)(70y)}$
$\implies \frac{N_3}{N_{\circ}}=0.185=18.5\%$
