Beta decay occurs when a neutron in an unstable nucleus decays into a proton and releases an electron and an antineutrino - Scottish Highers Physics - Question 7 - 2023

Element Z, corresponding to the atomic number of nitrogen in the reaction, is iodine (I) with the atomic number 53.

In the Standard Model, the type of fermion that includes electrons is known as 'leptons'.

The fundamental force associated with beta decay is the weak nuclear force.

The work done by the electric field on the proton can be calculated using the formula:

$W = QV$

Given that the charge of a proton is approximately $1.60 \times 10^{-19} C$ and the voltage is 32.0 kV, we can determine:

$W = (1.60 \times 10^{-19} C)(32.0 \times 10^{3} V) = 5.12 \times 10^{-15} J$

Using the work-energy principle, we know:

$W = \frac{1}{2}mv^2$

Substituting in the mass of a proton (approximately $1.67 \times 10^{-27} kg$), we get:

$5.12 \times 10^{-15} J = \frac{1}{2}(1.67 \times 10^{-27} kg) v^2$

Rearranging and solving for v gives:

$v = \sqrt{\frac{2(5.12 \times 10^{-15})}{1.67 \times 10^{-27}}} \approx 2.47 \times 10^6 m/s$.

An alternating voltage is used in the cyclotron to ensure that the electric field is always in the correct direction. This ensures that protons accelerate effectively and consistently as they cross the gap between the dees.

The direction of the force exerted on the proton can be determined using the right-hand rule. As the proton enters the magnetic field, it will experience a force directed 'up the page' due to the magnetic field applied perpendicular to its motion.