Question

Radium- 226 decays as ${ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He} .\( If the \){ }_{88}^{226}$ Ra nucleus is at rest before the decay and the \({ }_{86}^{222} \mathrm{Rn}\) nucleus is in its ground state, estimate the kinetic energy of the \(\alpha\) particle. (Assume that the \({ }_{86}^{222} \mathrm{Rn}\) nucleus takes away an insignificant fraction of the kinetic energy.)

Short answer

Question: Estimate the kinetic energy of an alpha particle when a Radium-226 nucleus decays and is initially at rest, assuming that the Radon-222 nucleus takes away an insignificant fraction of the kinetic energy. Answer: The kinetic energy of the alpha particle is approximately 4.870 MeV.

Step-by-step solution

Understand the conservation of momentum in nuclear reactions

In nuclear reactions, momentum is conserved. Since the Radium-226 nucleus is initially at rest, the combined momentum of the Radon-222 nucleus and the alpha particle after the decay should be equal to their combined momentum before the decay, which is zero.

Express the conservation of momentum

Let \(v_{Rn}\) be the velocity of Radon-222 nucleus and \(v_\alpha\) be the velocity of the alpha particle after the decay. We can write the conservation of momentum as follows: $$m_{Rn}v_{Rn}+m_\alpha v_\alpha = 0$$

Solve for the velocity of the alpha particle

We need to find the velocity of the alpha particle, \(v_\alpha\). By rearranging the equation from step 2, we get: $$v_\alpha = -\frac{m_{Rn}}{m_\alpha}v_{Rn}$$

Calculate the kinetic energy of the alpha particle

The kinetic energy (KE) of an object is given by the formula \(\frac{1}{2}mv^2\). Therefore, the kinetic energy of the alpha particle is: $$KE_\alpha = \frac{1}{2}m_\alpha v_\alpha^2 = \frac{1}{2}m_\alpha \left(-\frac{m_{Rn}}{m_\alpha}v_{Rn}\right)^2$$

Assume the Radon-222 nucleus takes away an insignificant fraction of the kinetic energy

We are given that the Radon-222 nucleus takes away an insignificant fraction of the kinetic energy. Therefore, we can approximate the kinetic energy of the alpha particle as the total kinetic energy released in the decay process. The sum of the kinetic energies can be written as: $$KE_\alpha + KE_{Rn} \approx KE_\alpha$$

Estimate the kinetic energy of the alpha particle

The decay energy of Radium-226 is approximately \(4.870\) MeV. We can estimate that most of this energy would be taken by the alpha particle. Hence, the kinetic energy of the alpha particle is approximately: $$KE_\alpha \approx 4.870\,\text{MeV}$$