State the principle of conservation of momentum - Leaving Cert Physics - Question 12 - 2016

During the alpha-decay of polonium-212, the daughter nucleus produced is lead-208 (Pb-208). This is due to the loss of an alpha particle, which consists of 2 protons and 2 neutrons from the original nucleus.

The kinetic energy (KE) of the emitted alpha-particle is given as 8.9 MeV. First, we convert this energy into joules:

$E = 8.9 ext{ MeV} = 8.9 imes 1.602 imes 10^{-13} ext{ J} = 1.426 imes 10^{-12} ext{ J}$

The relationship between kinetic energy and velocity is given by the equation:

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

Where:

• m is the mass of the alpha particle (approximately $4 imes 10^{-27}$ kg).
• Rearranging gives: $v = \sqrt{\frac{2E}{m}}$

Substituting the values:

$v = \sqrt{\frac{2 \times 1.426 \times 10^{-12} ext{ J}}{4 \times 10^{-27} ext{ kg}}} \approx 2.07 \times 10^7 ext{ m/s}$

Using conservation of momentum, since the initial momentum was zero:

$ext{Momentum before} = 0 = ext{Momentum after} = m_{alpha} v_{alpha} + m_{daughter} v_{daughter}$

Given that the mass ratio of lead-208 to the alpha particle is approximately 208:4, we have:

• Let the mass of the alpha particle be $m_{alpha} = 4m$ and the mass of the daughter nucleus be $m_{daughter} = 208m$.

Thus, substituting in: $0 = 4m imes v_{alpha} + 208m imes v_{daughter}$

This implies: $v_{daughter} = -\frac{4}{208} v_{alpha}$

Calculating it: $v_{daughter} = -\frac{4}{208} (2.07 \times 10^7) \approx -4.0 \times 10^{6} ext{ m/s}$

The negative sign indicates that the daughter nucleus moves in the opposite direction to the emitted alpha particle.