The Law of Radioactive Decay (DP IB Physics: HL): Revision Note

Strontium-90 decays with the emission of a β-particle to form Yttrium-90. The decay constant of strontium-90 is 0.025 year ^-1.

Determine the activity A of the sample after 5.0 years, expressing the answer as a fraction of the initial activity A₀.

Answer:

Step 1: Write out the known quantities

• Decay constant, λ = 0.025 year ^-1

• Time interval, t = 5.0 years

• Both quantities have the same unit, so there is no need for conversion

Step 2: Write the equation for activity in exponential form

$A=A_0{e}^{-\lambda t}$

Step 3: Rearrange the equation for the ratio between A and A₀

$\frac{A}{A_0}=e^{-\lambda t}$

Step 4: Calculate the ratio A/A₀

$\frac{A}{A_0}=e^{-\left(0.025\times 5\right)}=0.88$

• Therefore, the activity of strontium-90 decreases by a factor of 0.88, or 12%, after 5 years

A space probe uses a source containing 4.0 kg of plutonium-238.

Plutonium-238 is an alpha-emitter with a half-life of 87.7 years. Each alpha decay releases 5.5 MeV per emission. The space probe converts this into electrical energy with an efficiency of 32%.

The space probe can continue to operate as long as the power output is maintained at 0.4 kW or above.

Estimate the time, in years, the source is expected to supply power to the space probe.

Answer:

Step 1: List the known quantities

• Mass of Pu-238 = 4.0 kg = 4000 g

• Molar mass of Pu-238 = 238 g mol⁻¹

• Avogadro's constant, $N_A$ = 6.02 × 10²³ mol⁻¹

• Half-life of Pu-238 = 87.7 years

• Energy released per alpha decay = 5.5 MeV

• 1 electronvolt (eV) = 1.6 × 10⁻¹⁹ J

• Efficiency = 32% = 0.32

• Final power output, P = 0.4 kW = 400 W

Step 2: Calculate the initial number of nuclei present in the source

• 238 g of plutonium-238 contains 6.02 × 10²³ atoms (Avogadro's number), so in 4 kg:

Number of nuclei: $N=\frac{mass\times N_A}{molarmass}$

Initial number of nuclei: $N_0=\frac{4000\times \left(6.02\times {10}^{23}\right)}{238}=1.012\times {10}^{25}$ nuclei

Step 3: Calculate the initial activity of the source

Decay constant:  $\lambda =\frac{\ln 2}{t_{1/2}}$

Activity: $A=\lambda N$

• Combining these gives:

Initial activity: $A_0=\frac{N_0\ln 2}{t_{1/2}}$

$A_0=\frac{\left(1.012\times {10}^{25}\right)\times \ln 2}{87.7\times \left(24\times 60\times 60\times 365\right)}=2.54\times {10}^{15}$Bq

Step 4: Calculate the initial power output of the source

Power output: $P=\frac{∆E}{∆t}$

Energy released per decay: $E=\left(5.5\times {10}^6\right)\times \left(1.6\times {10}^{-19}\right)=8.8\times {10}^{-13}$ J

• Activity represents the decays per second, so:

Initial power output: $P_0=A_{0}E$

$P_0=\left(2.5\times {10}^{15}\right)\times \left(8.8\times {10}^{-13}\right)=2200$ W

• The electrical power transferred to the probe is:

$P_0=2200\times 0.32=704$ W

Step 5: Use the exponential decay equation to calculate the time of operation

• The power available is proportional to the activity of the isotope, so: 

Exponential decay of power: $P=P_0{e}^{-\lambda t}$

$\frac{P}{P_0}=e^{-\lambda t}$

$\ln \left(\frac{P}{P_0}\right)=-\lambda t$

$t=-\frac{1}{\lambda }\ln \left(\frac{P}{P_0}\right)=-\frac{t_{1/2}}{\ln 2}\ln \left(\frac{P}{P_0}\right)$

$t=-\frac{87.7}{\ln 2}\ln \left(\frac{400}{704}\right)=71.5$ years

• Therefore, the source is expected to supply power to the space probe for 71.5 years

The symbol e is used to represent the exponential constant and is approximately equal to e = 2.718

Make sure you are comfortable using the exponential function on your calculator, it is the button labelled $e^x$

The inverse function of $e^x$ is the natural logarithmic function, $\ln y$

The rules for exponential functions are the same as the rules for logarithmic functions, so, if $y=e^x$, then $x=\ln y$