Answer to Question #326139 in Differential Equations for nazir

Question

. The radioactive isotope carbon-10 has a half-life of 20 seconds.

a. How much time is required so that only 1/16 of the original amount remains?

b. Find the rate of decay at this time.

Answer 1

$T_{1/2}=20secs$

$\lambda=ln2/T_{1/2}$

a. $1 \rightarrow 1/2$ takes 20seconds

$1/2 \rightarrow 1/4$ takes 20 seconds

$1/4 \rightarrow 1/8$ takes 20 seconds

$1/8 \rightarrow 1/16$ takes 20 seconds

It takes 80secs for 1/16of the original to remain.

b.$decay\ rate=-\frac{dN}{dt}=\lambda N$

10g=6.03×10²³ molecules

decay rate=0.035×6.03×10^{23}=2.1×10^{22}0.035×6.03×10²³ =2.1×10²²

Particles per sec