E3-1. Isotopes, Background Radiation, and Radioactive Particles

Example 1: Deflection in a Uniform Magnetic Field (Classic IB Mechanics Bridge)

Problem: A radioactive source emits a mixed beam of alpha ($\alpha$), beta-minus ($\beta^-$), and gamma ($\gamma$) radiation. The beam passes through a velocity selector, so all particles enter a uniform magnetic field $B$ (directed into the page) at the exact same velocity $v$.
(a) Describe the direction of deflection for each of the three radiations.
(b) Analytically prove which particle ($\alpha$ or $\beta^-$) will travel in a circular path with the larger radius, and calculate the exact ratio of their radii ($r_{\alpha} / r_{\beta}$). Assume $m_{\alpha} \approx 4\text{ u}$ and $m_{\beta} \approx \frac{1}{1836}\text{ u}$.

Solution:

Part (a): Direction of Deflection:
Using Fleming's Left-Hand Rule (or Right-Hand Palm Rule):
• The $\alpha$ particle is positive. Current is in the direction of velocity. The magnetic force deflects it upwards.
• The $\beta^-$ particle is negative. Current is opposite to velocity. The magnetic force deflects it downwards.
• The $\gamma$ ray has no charge. It experiences no magnetic force and travels straight through undeflected.

Part (b): Radii of Curvature:
Equating magnetic force to centripetal force: $qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}$. Because $v$ and $B$ are constant for both, the radius is proportional to the mass-to-charge ratio ($m/q$).
• For Alpha: $q_{\alpha} = +2e$, $m_{\alpha} = 4\text{ u}$. Thus, $r_{\alpha} \propto \frac{4}{2} = 2$.
• For Beta: $q_{\beta} = -1e$, $m_{\beta} = \frac{1}{1836}\text{ u}$. Thus, $r_{\beta} \propto \frac{1/1836}{1} \approx 0.00054$.
To find the exact ratio: $$\frac{r_{\alpha}}{r_{\beta}} = \frac{\frac{m_{\alpha} v}{q_{\alpha} B}}{\frac{m_{\beta} v}{q_{\beta} B}} = \left(\frac{m_{\alpha}}{m_{\beta}}\right) \left(\frac{q_{\beta}}{q_{\alpha}}\right) = \left(\frac{4}{1/1836}\right) \left(\frac{1}{2}\right) = (7344) \times (0.5) = \mathbf{3672}$$ The alpha particle's radius is $3672$ times larger than the beta particle's. The beta particle curves very tightly, while the massive alpha particle barely curves at all!

Example 2: University Bridge — Statistics of Random Decay

Problem: Because radioactive decay is a purely random, stochastic process, the number of counts $N$ detected by a GM tube in a given time interval follows a Poisson distribution. In Poisson statistics, the absolute uncertainty (standard deviation) of a count is given by $\sigma = \sqrt{N}$.
A student measures the background radiation for 10 minutes and records 250 counts. They then place a radioactive source near the detector and record 3400 counts in 5 minutes. Calculate the corrected count rate in counts per minute (cpm) and determine its absolute uncertainty.

Solution:

Step 1: Calculate the background rate ($R_b$) and its uncertainty. $$N_b = 250 \implies \sigma_b = \sqrt{250} \approx 15.8 \text{ counts}$$ $$R_b = \frac{250}{10} = 25.0 \text{ cpm}$$ The uncertainty in the rate is $\frac{15.8}{10} = 1.58 \text{ cpm}$. So, $R_b = 25.0 \pm 1.6 \text{ cpm}$.

Step 2: Calculate the total source rate ($R_{total}$) and its uncertainty. $$N_{total} = 3400 \implies \sigma_{total} = \sqrt{3400} \approx 58.3 \text{ counts}$$ $$R_{total} = \frac{3400}{5} = 680 \text{ cpm}$$ The uncertainty in the rate is $\frac{58.3}{5} = 11.66 \text{ cpm}$. So, $R_{total} = 680 \pm 12 \text{ cpm}$.

Step 3: Calculate the corrected count rate ($R_c$) and propagate the errors. $$R_c = R_{total} - R_b = 680 - 25 = \mathbf{655 \text{ cpm}}$$ When adding or subtracting quantities with independent random uncertainties, we add their absolute uncertainties in quadrature (Pythagorean addition): $$\sigma_c = \sqrt{(\sigma_{R\_total})^2 + (\sigma_{R\_b})^2} = \sqrt{(11.66)^2 + (1.58)^2} = \sqrt{135.95 + 2.50} = \sqrt{138.45} \approx \mathbf{11.8 \text{ cpm}}$$ The final corrected count rate is strictly reported as $655 \pm 12 \text{ cpm}$.