The half-life of iodine-131 is 7.2 days. How long will it take for a sample of this substance to decay to 30% of its original amount?
Answers
half life = 7.2 days
Rate constant , K = ln(2)/ half-life = 0.693 / 7.2
we know that
ln(A/A0) = - Kt
=> t = - {ln(A/A0)}/K
= - {ln(0.3/1)} / {ln(2) / 7.2}
= 7.2 × ln(10/3) / ln(2)
= 7.2 × {log(10)-log(3)} / log(2)
= 7.2 × (1 - 0.477) / 0.301
= 7.2 × 0.523 / 0.301
= 12.51
so, the required time = 12.5 days
Step-by-step explanation:
It is given:
Charcoal has the initial activity, which is denoted as A_{0}=15.3A
0
=15.3 disintegrations per minute per gram.
Charcoal has the half-life, T 12=5730T12=5730 years
After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram
Constant of disintegration,
\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1λ=0.693T12=0.6935370y−1
For the action to attain 12.3 disintegrations per minute per gram, let the time taken by the sample at a time of t year.
Sample’s activity,
A=A O e-\lambda tA=AOe−λt
A=A O e-0.6935730 \times tA=AOe−0.6935730×t
\Rightarrow \mathrm{t}=1804.3 \text { years }⇒t=1804.3 years