Question

The half life of Pb-210 is 22 years. A decayed animal shows 16% of the original Pb-210 remains; how long has the animal been deceased to the nearest tenth of a year?
58.2 years


58.0 years


0.1 years


0.12 years

Answer 1

Answer:

The correct option is:   58.2 years.

Step-by-step explanation:

The half-life formula is:   $N(t)= N_{0}(\frac{1}{2})^\frac{t}{t_{1/2}}$ , where $N_{0}=$ Original amount, $N(t)=$ Final amount after $t$ years and $t_{1/2}=$ Half-life in years.

The half life of Pb-210 is 22 years. So,  $t_{1/2}= 22$ years.

A decayed animal shows 16% of the original Pb-210 remains. That means, if $N_{0}=100$, then $N(t)= 16$.

Plugging these values into the above formula, we will get......

$16= 100(\frac{1}{2})^\frac{t}{22}\\ \\ \frac{16}{100}= \frac{100(\frac{1}{2})^\frac{t}{22}}{100}\\ \\ 0.16=(\frac{1}{2})^\frac{t}{22}$

Taking logarithm on both sides.......

$log(0.16)=log[(\frac{1}{2})^\frac{t}{22}]\\ \\ log(0.16)=\frac{t}{22}log(\frac{1}{2})\\ \\ \frac{t}{22}=\frac{log(0.16)}{log(\frac{1}{2})}\\ \\ t= 22*\frac{log(0.16)}{log(\frac{1}{2})}=58.1648... \approx 58.2$

(Rounded to the nearest tenth)

So, the animal has been deceased for 58.2 years.