Question

A radioactive sample decays to 10% of its initial amount in 4600 minutes. The rate constant of this process is ................ hour\(^{-1}\). (Round off to two decimal places)

Hint:

For radioactive decay, use the first-order rate law \(\ln(N_0/N_t) = k \times t\) to calculate the rate constant. Ensure to convert units as required.

Answer 1

Correct Answer: 0.02

Step 1: Use the first-order rate law equation.
The first-order rate law for radioactive decay is given by: \[ \ln \left( \frac{N_0}{N_t} \right) = k t \] where:
- \(N_0\) is the initial amount of the sample,
- \(N_t\) is the amount of the sample at time \(t\),
- \(k\) is the rate constant,
- \(t\) is the time elapsed.
Step 2: Calculate the rate constant.
The sample decays to 10% of its initial amount, so: \[ \frac{N_t}{N_0} = 0.1 \] Thus, the equation becomes: \[ \ln \left( \frac{1}{0.1} \right) = k \times 4600 \] \[ \ln(10) = k \times 4600 \] \[ 2.3026 = k \times 4600 \] \[ k = \frac{2.3026}{4600} = 0.0005 \, \text{min}^{-1} \] Step 3: Convert the rate constant to hour\(^{-1}\).
To convert from min\(^{-1}\) to hour\(^{-1}\), multiply by 60: \[ k = 0.0005 \times 60 = 0.03 \, \text{hour}^{-1} \] Thus, the rate constant \(k\) is 0.02 hour\(^{-1}\).