# Compute the energy of the emitted alpha for Na20 decaying to an excited state

• llatosz
In summary, the excited state of 20Na decays to the ground state of 16O with an energy of 2.88634 MeV.
llatosz
I think I got this right, I just want a second opinion to know if my concepts are correct

1. Homework Statement

20Na decays to an excited state of 20Ne through the emission of positrons of maximum kinetic energy 5.55 MeV. The excited state decays by ##\alpha## emission to the ground state of 16O. Compute the energy of the emitted ##\alpha##

## Homework Equations

##T_{\alpha}=\frac{Q}{1+\frac{m_{\alpha}}{m_{X'}}}## where ##T_{\alpha}## is the kinetic energy of the alpha particle, ##m_{\alpha}## is the mass of that alpha particle, and ##m_{X'}## is the mass of the daughter nucleus.

##Q=\Delta mc^2##

## The Attempt at a Solution

For 20Na →20Ne,
##Q_0## = ##\Delta mc^2## = 13.8885 MeV. Of that 13.8885 MeV, 5.55 MeV goes to positrons so 13.8885-5.55 = 8.3385 MeV is left, which is the excitation energy.
For 20Ne*16O + 4He, where * denoted an excited state,
##Q_1## = 8.3385 + ##\Delta mc^2## = 8.3385 + (-4.72987252) = 3.6086295 MeV = ##Q_1##
##T_{\alpha} =\frac{Q_1}{1+\frac{m_{\alpha}}{m_{X'}}} = 2.88634 MeV = T_{\alpha}##

Now this would be the kinetic energy. Would I have to take the square root of the kinetic energy squared plus the rest mass energy squared? Or is the 2.88 my answer? I think it is the 2.88 because it should not be relativistic so rest mass energy should not matter here in this case. What are your thoughts?

I'm quite sure the question asks for the kinetic energy of the alpha particle, not the total energy.

mfb said:
I'm quite sure the question asks for the kinetic energy of the alpha particle, not the total energy.
Nah, it just says "energy"

Yes, and I'm quite sure that refers to the kinetic energy.

mfb said:
Yes, and I'm quite sure that refers to the kinetic energy.
Oh okay. So in that case do you think I did it correctly with 2.88 MeV as my answer?

I didn't check the numbers, the approach is right.

Okay great, thank you!

## 1. What is the process of alpha decay?

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, consisting of two protons and two neutrons, from its nucleus.

## 2. How is the energy of the emitted alpha particle calculated?

The energy of the emitted alpha particle is calculated using the equation E=mc², where E is energy, m is mass, and c is the speed of light. The mass of the alpha particle is subtracted from the initial mass of the parent nucleus to determine the energy released.

## 3. What is the energy level of the excited state in this case?

The energy level of the excited state in this case refers to the energy difference between the ground state and the excited state of the nucleus. This energy level can be calculated using the Bohr model or other quantum mechanical models.

## 4. How does the energy of the emitted alpha particle affect the stability of the nucleus?

The energy of the emitted alpha particle plays a crucial role in determining the stability of the nucleus. If the energy released during alpha decay is too high, it can cause the nucleus to become unstable and undergo further decay processes.

## 5. Is the energy of the emitted alpha particle constant for all alpha decays?

No, the energy of the emitted alpha particle can vary for different alpha decay processes. It depends on factors such as the initial mass of the parent nucleus and the energy level of the excited state. However, the energy released in alpha decay is always a fixed amount for a particular nucleus.

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