Nuclear Physics: Question 9

Worked solution

Part (a): Why background must be subtracted

The Geiger-Muller tube detects ionising radiation from all sources in the vicinity, not just X. Cosmic rays and naturally-occurring radioactive material in the surroundings produce a small, roughly constant background count rate that is present whether or not X is in the room. This background does not decay with time, so if it is left in the readings, the measured count rate will not follow the clean exponential decay law expected of X alone. Subtracting it isolates the count rate due only to the decay of X.

Part (b): Corrected count rates

Subtracting the background count rate of $20$ counts per minute from each measured value: $C_0=620-20=600\text{ counts/min (at }t=0)$ $C_{15}=170-20=150\text{ counts/min (at }t=15.0\text{ min)}$

$C_0=\boxed{600}\text{ counts/min},\qquad C_{15}=\boxed{150}\text{ counts/min}$

Part (c): Show that the half-life is 7.50 minutes

The corrected count rate has fallen by a factor of: $\frac{C_{15}}{C_0}=\frac{150}{600}=\frac{1}{4}$

Since $\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2$, the count rate has halved twice in the $15.0$ minutes elapsed. That is, exactly two half-lives have passed: $2\,t_{1/2}=15.0\text{ min}\implies t_{1/2}=\frac{15.0}{2}=7.50\text{ min}$

Recompute as a check, using the exponential decay law directly: $C=C_0e^{-\lambda t}$ gives $0.25=e^{-15.0\lambda}$, so $\lambda=\dfrac{\ln4}{15.0}=\dfrac{1.3863}{15.0}=0.09242\text{ min}^{-1}$, and hence $t_{1/2}=\dfrac{\ln2}{\lambda}=\dfrac{0.6931}{0.09242}=7.50\text{ min}$, the same result.

So $t_{1/2}\approx\boxed{7.50}$ minutes, as required to show.

Part (d): Decay constant in per second

Converting the half-life to seconds: $t_{1/2}=7.50\times60=450\text{ s}$

Using $\lambda=\ln2/t_{1/2}$: $\lambda=\frac{0.6931}{450}=1.540\times10^{-3}\text{ s}^{-1}$

Recompute as a check, working in minutes first and converting afterwards: $\lambda=\dfrac{\ln2}{7.50}=0.09242\text{ min}^{-1}$, and dividing by $60\text{ s/min}$ gives $\dfrac{0.09242}{60}=1.540\times10^{-3}\text{ s}^{-1}$, the same result.

$\lambda=\boxed{1.54\times10^{-3}}\text{ s}^{-1}\text{ (3 s.f.)}$

Final answers

• (a) Background radiation does not come from X and does not decay, so it must be subtracted to isolate X’s exponential decay
• (b) $C_0=\boxed{600}$ counts/min, $C_{15}=\boxed{150}$ counts/min
• (c) $t_{1/2}\approx\boxed{7.50}$ minutes
• (d) $\lambda=\boxed{1.54\times10^{-3}}\text{ s}^{-1}$