Minimum Energy Of A Beta Particle

[kkrizka]

Homework Statement

When Ne (Z=10) decays to Na (Z=11), what is the maximum kinetic energy of the emitted electron? What is the minimum energy? What is the energy of the neutrino in each case? Ignore recoil of the daughter nucleus.

Homework Equations

$$E = mc^2$$

The Attempt at a Solution

The following is the decay:
$$^{23}_{10} Ne \rightarrow ^{23}_{11} Na + \beta ^- + n + Q$$

I know that there will be extra energy, Q, emitted in the decay and I know I can find it using the difference in mass between the products and initial particles. Now I'm assuming that this energy will go towards the kinetic energy of the neutron and the electron. But first, the momentum should be conserved. So:
$$m_{\beta} v_{\beta} = m_n v_n$$ or $$v_n = \frac{m_{\beta}}{m_n} v_{\beta}$$

Then I can use this together with the energy equation:
$$\frac{m_{\beta} v_{\beta}^2}{2} + \frac{m_n v_n^2}{2} = Q$$
$$\frac{m_{\beta} v_{\beta}^2}{2} + \frac{m_{\beta}^2 v_{\beta}^2}{ 2 m_n } = Q$$
$$\frac{m_{\beta} v_{\beta}^2}{2} = \frac{Q}{( 1 + \frac{m_{\beta}}{m_n} )}$$

So as the mass of the neutrino approaches 0, then the fraction m_\beta / m_n approaches infinity (so 1+ that approaches infinity) and Q divided by a very big number would be about 0.

That is what am I unsure of, is the kinetic energy of the electron really 0 or did I make a mistake somewhere in my assumptions?

My second question concerns the minimum kinetic energy of the electron. Would the minimum KE be when all of Q is heat, so in that case KE=0?

 

[jbsteele]

Or would the KE be greater than 0?Finally, what is the energy of the neutrino in each case?Thank you for any help!

 

[baba89]

Or would there always be some minimum energy required to overcome the binding energy of the nucleus and release the electron?

I would like to clarify a few points about the concept of minimum energy of a beta particle. First, it is important to understand that the energy of a beta particle is not fixed, but rather has a range of values. This range is determined by the conservation of energy and momentum in the decay process, as you have correctly stated in your solution attempt.

However, the minimum energy of a beta particle is not 0. This is because there is always a minimum amount of energy required to overcome the binding energy of the nucleus and release the beta particle. This energy is known as the Q-value of the decay, and as you have correctly pointed out, it is determined by the difference in mass between the initial and final particles.

In the case of the Ne (Z=10) decaying to Na (Z=11), the Q-value can be calculated as follows:
Q = (m_{Ne}-m_{Na}-m_{e})c^2
= (23.0179 u - 23.0059 u - 0.000549 u) (931.5 MeV/u)
= 0.01135 u (931.5 MeV/u)
= 10.56 MeV

This means that the minimum energy of the beta particle in this decay process is 10.56 MeV. However, as you have correctly stated, the maximum kinetic energy of the beta particle is not equal to the Q-value. This is because some of the energy is also given to the neutrino in the decay process.

In conclusion, the minimum energy of a beta particle is not 0, but rather is determined by the Q-value of the decay process. The maximum kinetic energy of the beta particle is also not equal to the Q-value, but rather is determined by the conservation of energy and momentum in the decay process. The energy of the neutrino in each case can be calculated by subtracting the kinetic energy of the beta particle from the Q-value. I hope this clarifies any confusion and helps with your solution.