Figure 11 shows radioactive decay of one nuclide can be modelled by draining water through a tap from a cylindrical tube - AQA - A-Level Physics - Question 6 - 2018 - Paper 2

To represent a radioactive sample with a greater number of nuclei, we can increase the volume of water used in the cylindrical tube. This allows for a greater number of water molecules to simulate the behavior of a larger sample of the nuclide.

To represent a radioactive sample of a nuclide with a smaller decay constant, we should use a narrower tube or a smaller capillary. This will decrease the rate of flow, simulating a slower decay process.

The half-life ($T_{1/2}$) can be calculated using the formula:

$T_{1/2} = \frac{0.693}{\lambda}$

where $\lambda = 1.42 \times 10^{-11} year^{-1}$. Thus, we calculate:

$T_{1/2} = \frac{0.693}{1.42 \times 10^{-11}} \approx 4.87 \times 10^{10} years.$

Therefore, the half-life of $^{87}Rb$ is approximately 4.87 x 10^10 years.

We can use the decay formula:

$N = N_0 e^{-\lambda t}$

Where:

• $N$ = final amount (1.23 mg)
• $N_0$ = initial amount
• $\lambda = 1.42 \times 10^{-11} year^{-1}$
• $t = 4.47 \times 10^9 years$

First, we rearrange to find $N_0$:

$N_0 = N e^{\lambda t}$

Calculating this gives:

$N_0 = 1.23 mg \times e^{1.42 \times 10^{-11} \times 4.47 \times 10^9}.$

Calculating the exponent gives us approximately:

$N_0 \approx 1.23 mg \times 1.063 \approx 1.31 mg.$

Thus, the initial mass of $^{87}Rb$ was approximately 1.31 mg.

The activity (A) can be calculated using the formula:

$A = \lambda N$

where:

• $\lambda = 1.42 \times 10^{-11} year^{-1}$
• $N$ = number of nuclei, which can be calculated using:

$N = \frac{mass}{mass_{nucleus}}$

The mass of one $^{87}Rb$ nucleus = 1.44 x 10^-25 kg. Thus, for the sample:

$N = \frac{1.23 \times 10^{-3} g}{1.44 \times 10^{-25} kg} \approx 8.54 \times 10^{21} nuclei.$

Finally, substituting into activity:

$A = (1.42 \times 10^{-11}) (8.54 \times 10^{21}) \approx 121.07 \ mBq.$

Therefore, the activity is approximately 121.07 mBq.