Lise Meitner and Marie Curie are the only women scientists to have elements named after them - Leaving Cert Physics - Question 9 - 2016

Radioactivity refers to the spontaneous disintegration of a nucleus, which results in the emission of radiation. This decay process can release particles such as alpha and beta particles or gamma rays, contributing to the transformation of the original nucleus into different elements or isotopes.

To calculate the energy released, we can use the mass-energy equivalence principle, given by the formula: $E = mc^2$ Firstly, we compute the mass before and after the fission, using the provided mass values:

• Mass before: $3.9696 \times 10^{-28}$ kg
• Mass after: $2.9 \times 10^{-28}$ kg Mass defect: $\Delta m = m_{before} - m_{after}$ Substituting the values gives us: $\Delta m = (3.9696 \times 10^{-28} - 2.9 \times 10^{-28}) \text{ kg} = 1.0696 \times 10^{-28} \text{ kg}$ Now we calculate the energy released: $E = \Delta m \times c^2 \approx (1.0696 \times 10^{-28}) \text{ kg} \times (3 \times 10^8)^2 \approx (1.0696 \times 10^{-28}) \times (9 \times 10^{16}) \text{ J} \approx 9.67 \times 10^{-12} \text{ J}$

In a fission chain reaction, typically 1 neutron is required to sustain the reaction. This means that on average, each fission event will release enough neutrons to continue the process. However, to ensure a self-sustaining reaction, the relationship is often expressed as:

• 1: uncontrolled reaction

• <1: chain-reaction ending Thus, for the reaction to remain safe and self-sustaining, it needs to balance around 1 neutron per fission.

The function of the moderator in a fission reaction is to slow down the emitted neutrons. By reducing the kinetic energy of fast neutrons, moderators enhance the likelihood of these neutrons being captured by other fissile nuclei, allowing for more fission events to occur. This makes the reaction more efficient and manageable.

The nuclear equation for the decay of radium-225 ($_{88}^{225}Ra$) to actinium ($_{89}^{225}Ac$) is as follows:
$_{88}^{225}Ra \rightarrow _{89}^{225}Ac + _{0}^{1}e - \text{ (Beta emission)}$

The relationship between activity (A), number of particles (N), and decay constant (λ) is given by the formula:
$A = \lambda N$ To find N, we first need to determine λ. The decay constant λ is calculated using the half-life (T_1/2) formula:
$\lambda = \frac{ln(2)}{T_{1/2}} \approx \frac{0.693}{14.9 \times 24 \times 60 \times 60} \approx 8.2 \times 10^{-6} s^{-1}$
Now substituting back into the activity equation gives: $N = \frac{A}{\lambda} = \frac{5600 Bq}{8.2 \times 10^{-6} s^{-1}} \approx 6.83 \times 10^{11} nuclei$.