Chapter 30, Problem 1OQ

To determine

The weight of ${}_{\text{27}}{}^{\text{60}}\text{C}\text{O}$ left after 10.6 years

Answer to Problem 1OQ

Solution:

b)20µg

Explanation of Solution

Given:

Half -life of ${}_{\text{27}}{}^{\text{60}}\text{C}\text{O}$, t_1/2 = 5.3 years

Initial weight of ${}_{\text{27}}{}^{\text{60}}\text{C}\text{O}$, N₀= 80µg

Formula used:

Half life of radioactive substance is given by,

${\text{t}}_{1/2}\text{ = }\frac{0.693}{\lambda }$ ……….. (eq.1)

Where,

λ→ Decay constant

Amount of radioactive substance left after a time ‘t’ is given by,

$\text{N = }{N}_0{e}^{-\lambda t}$ ……….. (eq.2)

Where,

N₀→ Initial amount

λ → Decay constant

Calculation:

From eqn.1 decay constant of ${}_{\text{27}}{}^{\text{60}}\text{C}\text{O}$ can be found as,

$\begin{array}{l}\lambda \text{ = }\frac{0.693}{t_{1/2}}\\ \text{ = }\frac{\text{0}\text{.693}}{5.3}\\ \text{ = 0}{\text{.1307 year}}^{-1}\end{array}$

Now from eqn.2, amount of ${}_{\text{27}}{}^{\text{60}}\text{C}\text{O}$ left after 10.6 years can be calculated

$\begin{array}{l}\text{N = 80}\times e^{-0.1307\times 10.6}\\ \text{ = 20}\text{.017 }\mu \text{g}\end{array}$