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

$T_{1 \over 2} =20secs$

$\lambda ={ln2 \over T_{1 \over 2}}= {0.693 \over 20}=0.035sec^{-1}$

a) $1\to{1 \over 2}$ takes 20secs

${1 \over 2} \to{1 \over 4}$ takes 20secs

${1 \over 4} \to{1 \over 8}$ takes 20secs

${1 \over 8} \to{1 \over 16}$ takes 20sec

$\therefore$ It takes 80secs for ${1 \over 16}$ of the original to remain

b)decay rate= $-{dN \over dt} =\lambda N$

$10g= 6.03×10^{23} molecules$

decay rate=$0.035×6.03×10^{23}=2.1×10^{22}$ Particles per sec