A nucleus of polonium $^{210}_{84}Po$ may decay to the stable isotope of lead $^{208}_{82}Pb$ through a chain of emissions following the sequence $eta^-$ decay - AQA - A-Level Physics - Question 7 - 2018 - Paper 2

More neutrons are needed to help bind protons together and increase binding energy, thus overcoming repulsion from the electromagnetic force.

$^{210}_{84}Po + e^- \rightarrow ^{210}_{83}Tl + u_e$

The sources of electromagnetic radiation emitted from the excited thallium nucleus can be explained as follows:

1. Decay Transitions: Following electron capture, the newly formed thallium nucleus may be in an excited state. It releases energy in the form of gamma radiation as it transitions to a lower energy state, occurring within the nucleus itself.

2. Relaxation of Atomic Electron States: When the thallium atom is initially formed in an excited state, its electrons may also drop from higher energy levels to lower ones, emitting X-rays as they transition within the electron shells around the nucleus.

Both processes result in the emission of significant electromagnetic radiation.

The metastable form of technetium $^{99m}_{43}Tc$ is ideal for medical diagnosis due to its favorable properties:

1. Short Half-Life: $^{99m}Tc$ has a short half-life of about 6 hours, meaning it decays quickly, reducing patient exposure to radiation while still providing clear imaging results.

2. Gamma Emission: It emits gamma rays, which can penetrate the body's tissues and are easily detectable by imaging technologies, making it effective for diagnostic imaging procedures without excessive damaging radiation.

3. Chemical Versatility: $^{99m}Tc$ can form various compounds, which can target different organs, allowing for tailored imaging based on specific medical needs.

These attributes make $^{99m}_{43}Tc$ a highly effective and suitable radioactive source for medical diagnostics.