Determine how much energy the alpha particle carries

[Woopa]

Homework Statement

Calculate the ratio of the kinetic energies of the alpha particle to that of the radon nucleus in the decay of radium (226 88Ra) to radon (222 86Rn). Assume that the radium nucleus decays at rest. Determine how much energy the alpha particle carries.

Relevant Equations

E=p^2/m

I can follow through all of this worked example until the final step 55/56x4.84= 4.75MeV

Where does the 56 come from?

 

[DrClaude]

Where does the 56 come from?

$$E_\mathrm{total} = E_\alpha + E_\mathrm{Rn}$$
so what is the relative energy of the α particle?

 

[Woopa]

$$E_\mathrm{total} = E_\alpha + E_\mathrm{Rn}$$
so what is the relative energy of the α particle?

the relative energy would be the ratio of 4: 222= 0.018?

 

[DrClaude]

the relative energy would be the ratio of 4: 222= 0.018?

That gives you the factor of 55.

What I meant is the ratio if the energy of the α particle to the total energy.

 

[Woopa]

Okay I think I've got it!

So energy of alpha particle 2(1.0073x931.55)+2(1.0087x931.55)=3756MeV
Total energy = 88x(1.0073x931.55)=138(1.0087x931.55)=212247Mev

Ratio total energy:alpha particle = 212247/3756=56

So the alpha particle makes up 1/56th of the total energy.

Why do we only think about the alpha particle and not include the energy of the radon nucleus in the ratio?

 

[PeroK]

Okay I think I've got it!

So energy of alpha particle 2(1.0073x931.55)+2(1.0087x931.55)=3756MeV
Total energy = 88x(1.0073x931.55)=138(1.0087x931.55)=212247Mev

Where are you getting these numbers from?

Ratio total energy:alpha particle = 212247/3756=56

That's not right. The answer should be $56:55$, as above.

So the alpha particle makes up 1/56th of the total energy.

No. It makes up 55/56 of the total kinetic energy.

 

[Steve4Physics]

Hi @Woopa.

$\frac {222}{4}$ has been incorrectly rounded to $55$. Note $\frac {222}{4} = 55.5$!

And $\frac {226}{4}$ has (in hidden/implied working) been incorrectly rounded to give the mystery value $56$. But $\frac {226}{4} = 56.5$.

This means $\frac {55}{56}$ should $\frac {55.5}{56.5}$. Or even better $\frac {222}{226}$

So the official answer leads something to be desired!

The last part is really just a simple algebra exercise. Suppose you have to divide a cake into two portions (A and B) so that A is ($\frac {222}{4}=)$ 55.5 times bigger than B. What proportion of the cake is A?

 

[neilparker62]

Where are you getting these numbers from?

Looks like the op is converting amu to electron volts for 2 proton and 2 neutrons (alpha particle) and 88 protons + 88 138 neutrons (radon).

 

[PeroK]

Looks like the op is converting amu to electron volts for 2 proton and 2 neutrons (alpha particle) and 88 protons + 88 neutrons (radon).

Ah, so converting the masses to MeV and then confusing mass-energy with kinetic energy.

 

[neilparker62]

Ah, so converting the masses to MeV and then confusing mass-energy with kinetic energy.

Yes. I think we have a lot of careful 'unravelling' to do here. Not least because the ratio $E_t:E_{\alpha}$ obtained in his calculation 212247:3756 is 56.5 : 1 which is more or less correct given the rounding. Right answer , wrong thinking always quite tricky to explain!

 

[neilparker62]

I found this reference useful. It explains that the decayed particles have a "mass deficit" which produces the particles' KE.

https://www.schoolphysics.co.uk/age...r structure/text/Nuclear_stability/index.html

 

[neilparker62]

Correction Radon 226 has 88 protons and 138 neutrons , not 88 neutrons!

 

[PeroK]

I found this reference useful. It explains that the decayed particles have a "mass deficit" which produces the particles' KE.

https://www.schoolphysics.co.uk/age16-19/Nuclear physics/Nuclear structure/text/Nuclear_stability/index.html

For the puposes of this question:

1) We have a mass deficit of $0.0052 \ u = 4.84 \ MeV$, which is not immediately apparent from the mass numbers, but requires a more detailed analysis of the mass-energy of each particle.

2) That energy is converted to the total KE (Kinetic Energy) of the two particles.

3) By analysing the relationship between momentum, mass and KE, we see that the KE is in inverse proportion to the mass. I.e. the lighter particle takes away most of the KE.

4) This leads directly to the ratio of KE being expressed in terms of the inverse ratio of the masses.

@Woopa when you followed the worked example, this is what you should have understood from it.

 

[Steve4Physics]

@PeroK beat me to it in Post #13. But since I've already drafted it...

Note the working in the official answer (given in Post#1) is wrong due to incorrect rounding:
55 should be 55.5 (=222/4)
56 should be 56.5 (=226/4)

The OP has (totally unnecessarily) found the rest-energies of the radium nucleus and the alpha particle by adding the rest-energies of their nucleons, giving:
Ra-226: 212247Mev
He-4: 3756MeV

In fact this calculation is (slightly) wrong because binding energies have been ignored. (And the error will equal the kinetic energy acquired by the decay fragements.)

It is to be expected that 212247:3756 ≈ 56.5:1 because $m_n≈m_p$ and 56.5 is the ratio of the particles’ nucleon-numbers: $\frac {226}{4} =56.5$.

Presumably the OP was given the total energy released (4.84MeV in the Post #1 official answer).

The OP’s original question is simply about how the fraction $\frac {55}{56} \times 4.84$ arises. In fact it should be $\frac {55.5}{56.5} \times 4.84$. See also my Post #7.

Edit - minor rewording

 

[Woopa]

Thanks for the clarification everyone!