. 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.

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T1/2=20secsT_{1/2}=20secsT1/2​=20secs

λ=ln2/T1/2lambda=ln2/T_{1/2}λ=ln2/T1/2​

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

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

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

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

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

b.decay rate=−dNdt=λNdecay rate=-frac{dN}{dt}=lambda Ndecay rate=−dtdN​=λN

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