Question

Assuming 1μg of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ × 10–1μg.
[Given : ln 10 = 2.303; log 2 = 0.30]

Answer 1

$kt=ln\frac {1}{1−x}$

$\frac {0.693}{30}(100)=ln\frac {⁡1}{1−x}$

$2.303=2.303\ log⁡\frac {1}{1−x}$
$⇒\frac {1}{1−x}=10$
$⇒1=10–10X$
$⇒X=\frac {9}{10}$
$⇒X =0.9$ μg
Amount of X remaining$=1–X$
$=1–0.9$
$=0.1$ μg
$=1×10^{−1}$ μg

So, the answer is $1$.