There are about a trillion neutrinos from the Sun passing through your hand every second - Leaving Cert Physics - Question 10 - 2015

The two other leptons are the muon and the tau.

One fundamental particle that is subject to the strong nuclear force is a quark.

Pauli proposed that a neutrino is emitted during beta-decay to account for the conservation of momentum and energy in the process, because the decay produces a proton and an electron, and a neutrino is necessary to ensure these quantities are conserved.

The nuclear equation for beta-decay is:

To calculate the energy released during beta-decay, we first determine the loss in mass:

• Mass loss: $1.395 \times 10^{-30} \text{ kg}$

Using Einstein’s equation, $E = mc^2$:

1. $E = (1.395 \times 10^{-30} \text{ kg}) \times (3.00 \times 10^8 ext{ m s}^{-1})^2$
2. $E = 1.25 \times 10^{-13} \text{ J}$
3. Converting to MeV: $E = 0.78 \text{ MeV}$.

It is much more difficult to detect a neutrino because neutrinos have no electric charge and interact very weakly with matter. They can pass through normal matter—including entire planets—almost without any interaction, making them nearly undetectable.

To calculate the radius of the circular path of the electron in the magnetic field, we use the formula:
$r = \frac{mv}{Bq}$
Where:

• $m$ is the mass of the electron: $9.11 \times 10^{-31} \text{ kg}$
• $v$ is the speed of the electron: $1.45 \times 10^6 \text{ m s}^{-1}$
• $B$ is the magnetic flux density: $90 \text{ mT} = 0.09 \text{ T}$
• $q$ is the charge of the electron: $1.60 \times 10^{-19} \text{ C}$

Calculating the radius:
$r = \frac{(9.11 \times 10^{-31} \text{ kg})(1.45 \times 10^6 \text{ m s}^{-1})}{(0.09 \text{ T})(1.60 \times 10^{-19} \text{ C})} \approx 0.247 \text{ m}$

A neutrino travels in a straight path in the magnetic field because it is neutral and therefore does not experience any Lorentz force. Unlike charged particles, neutrinos do not curve in a magnetic field and move in linear trajectories.