Question

Given the initial weight of 1 mg of radioactive \( ^{60} \)Co (half-life = 5.27 years), the amount disintegrated in 1 year (rounded off to two decimal places) is \(\underline{\hspace{2cm}}\) mg.

Hint:

To calculate the amount of a substance disintegrated in radioactive decay, use the decay constant derived from the half-life.

Answer 1

Correct Answer: 0.11

The amount of substance remaining after time \( t \) in a radioactive decay process is given by:
\[ N(t) = N_0 e^{-\lambda t} \] where \( \lambda \) is the decay constant and \( t \) is the time in years. The half-life \( t_{1/2} \) is related to \( \lambda \) by:
\[ \lambda = \frac{\ln 2}{t_{1/2}}. \] For \( ^{60} \)Co, \( t_{1/2} = 5.27 \) years, so:
\[ \lambda = \frac{\ln 2}{5.27} = 0.131 \, \text{per year}. \] The initial amount is \( N_0 = 1 \, \text{mg} \). The amount remaining after 1 year is:
\[ N(1) = 1 e^{-0.131 \times 1} = 0.877 \, \text{mg}. \] The amount disintegrated is:
\[ 1 - 0.877 = 0.123 \, \text{mg}. \] Thus, the amount disintegrated in 1 year is \( 0.12 \, \text{mg} \).