Energy of the photon emitted in a gamma ray decay

[Rahulrj]

Homework Statement

An Fe nucleus (A=57) decays from an excited stated to the ground state by emitting a gamma ray. The energy of the photon is 14.4 KeV when the nucleus is held fixed. If the nucleus is free to recoil then the energy of the photon emitted will be?

Homework Equations

$E = mc^2$
$E^2 = (pc)^2 + E_{0}^2$
$KE (T) = p^2/2m$

The Attempt at a Solution

So writing down the energy conservation
(E* - energy of Fe before decay, E' - energy of the lower state nucleus, T_{recoil} - KE from the recoiling of the nucleus)
Case 1: when nucleus is fixed $T_{recoil} = 0$
therefore
$E* = E'+14.4$
Case 2: When nucleus is allowed to recoil, $p_{photon} = p_{recoil}$
$E* = E'+E_{photon}+T_{recoil}$
Now $T_{recoil} = p_{recoil}^2/2m$ ( the KE of recoil is small enough that non relativistic equation can be used)
Substituting the equation in case 1
$E'+14.4= E'+E_{photon}+T_{recoil}$
$E_{photon}+T_{recoil} = 14.4$
I am stuck here as I do not know how to figure out T_{recoil}, Can somebody please help?

 

[kuruman]

How about momentum conservation?

 

[Rahulrj]

How about momentum conservation?

Yes I did consider momentum conservation However, I do not know how to go beyond $p_{photon} = p_{recoil}$
because I do not know the energy of photon or the nucleus at recoil which is the second case.

 

[kuruman]

You know that $p_{\gamma}=p_{nucleus}$ and you are looking for $E_{\gamma}$. Express $T_{recoil}$ first in terms of $p_{nucleus}$ and then in terms of $E_{\gamma}$.

 

[Rahulrj]

You know that $p_{\gamma}=p_{nucleus}$ and you are looking for $E_{\gamma}$. Express $T_{recoil}$ first in terms of $p_{nucleus}$ and then in terms of $E_{\gamma}$.

Yes that's what I did think of but the question got me confused, the $E_\gamma$ given in the question is for the case in which nucleus is fixed. So when I equate $p_\gamma = p_{recoil}$ isn't the $p_\gamma$ representing the momentum of photon for the case in which nucleus is free to recoil? isn't the momentum conservation representing the same case?

 

[Rahulrj]

I think I am getting the right answer with that substitution, this is how I did:
$E_{photon}+T_{recoil} = 14.4$
$T_{recoil} =p_{recoil}^2/2m$
$p_{recoil} = p_{photon}$
$T_{recoil} =p_{photon}^2/2m$
$T_{recoil} =E_{\gamma}^2/2mc^2$
mc^2 represents the energy of Fe nucleus = 51.3GeV (given A = 57)
$E_{photon}+E_{\gamma}^2/2mc^2 = 14.4$
$E_{photon}+207.36*10^6/102.6*10^9 = 14.4$
$E_{photon}=14.4 * 10^3 -2.021*10^{-3}$ and this is the option b.
the options: a) lower by approx 2kev
b) lower by approx 2mev
c) higher by approx 2ev
d) higher by approx 20ev
e) lower by approx 20mev
However my doubt still remains about the substitution of $E_\gamma$. The momentum equation was for the second case where the nucleus recoils. how is it right to substitute E of the photon in the first case to the momentum defined by the second case?

 

[kuruman]

$E_{photon}$ and $E_{\gamma}$ are one and the same thing. This is a process in which a single photon is emitted and we are looking for its energy with or without nuclear recoil.

Also, I don't understand where the 207.36*10⁶ came from in the equation $E_{photon}+207.36*10^6/102.6*10^9 = 14.4$. I suggest that you solve for the photon energy symbolically first and then put in the numbers. In other words, first find an algebraic expression for $E_{photon}$ (or $E_{\gamma}$) and then substitute. Use $E_{excited} - E_{ground}$ instead of 14.4 keV. This should help you see what's going on.

 

[Rahulrj]

$E_{photon}$ and $E_{\gamma}$ are one and the same thing. This is a process in which a single photon is emitted and we are looking for its energy with or without nuclear recoil.

Also, I don't understand where the 207.36*10⁶ came from in the equation $E_{photon}+207.36*10^6/102.6*10^9 = 14.4$. I suggest that you solve for the photon energy symbolically first and then put in the numbers. In other words, first find an algebraic expression for $E_{photon}$ (or $E_{\gamma}$) and then substitute. Use $E_{excited} - E_{ground}$ instead of 14.4 keV. This should help you see what's going on.

I took $E_{photon}$ to be the energy of photon when nucleus recoils and $E_{\gamma}$ to be energy of photon while nucleus remains fixed which is given to be 14.4 kev. So squaring $E_{\gamma}$ gives that value.

 

[kuruman]

However my doubt still remains about the substitution of $E_\gamma$.

Then you cannot substitute and your doubts are correct. You need to solve the equation with one $E_{\gamma}.$ It can be done.

 

[Rahulrj]

Then you cannot substitute and your doubts are correct. You need to solve the equation with one $E_{\gamma}.$ It can be done.

I do not get why you keep mentioning that there is only one E for photon. Clearly, the question is about a single photon having two values of E for two different situations (nucleus fixed and free to recoil) so in that case why or how would $E_{\gamma}$ $E_{photon}$ be the same?

and to solve with one E, in this equation $E_{photon}+E_{\gamma}^2/2mc^2 = 14.4$ I have to then replace $E_{\gamma}$ with $E_{photon}$? which would then become a quadratic equation and is that how it should be?

 

[kuruman]

The energy balance equation is always $E_{excited} = E_{ground} + E_{recoil} +E_{\gamma}$. Now $E_{excited} - E_{ground} = 14.4~keV$ regardless of whether the photon is emitted recoil-free or not. When recoil free, $E_{\gamma} = 14.4~ keV$ because $E_{recoil}=0$ . When not recoil-free, the problem is asking you to find $E_{\gamma}$ which will be less than $E_{excited} - E_{ground}$.

As long as you know what it stands for, you can call the photon's energy $E_{photon}$ if you wish or you can call it "Fred", it's just a placeholder. Like you say, the photon may have two energies depending on what process you are talking about. I prefer to use the symbol, $E_{\gamma}$ whenever I wish to represent a photon's energy when it is in the gamma ray regime.

 

[Rahulrj]

The energy balance equation is always $E_{excited} = E_{ground} + E_{recoil} +E_{\gamma}$. Now $E_{excited} - E_{ground} = 14.4~keV$ regardless of whether the photon is emitted recoil-free or not. When recoil free, $E_{\gamma} = 14.4~ keV$ because $E_{recoil}=0$ . When not recoil-free, the problem is asking you to find $E_{\gamma}$ which will be less than $E_{excited} - E_{ground}$.

As long as you know what it stands for, you can call the photon's energy $E_{photon}$ if you wish or you can call it "Fred", it's just a placeholder. Like you say, the photon may have two energies depending on what process you are talking about. I prefer to use the symbol, $E_{\gamma}$ whenever I wish to represent a photon's energy when it is in the gamma ray regime.

I too have written the same equations although with different representations in the first post itself. Now since you said I need to solve it using one $E_{\gamma}$ . The equation then turns out to be quadratic as I mentioned in previous post,

and to solve with one E, in this equation $E_{photon}+E_{\gamma}^2/2mc^2 = 14.4$ I have to then replace $E_{\gamma}$ with $E_{photon}$?

I don't think this is correct. It has to be either this way or the way I did before unless there is some other way of crude substitution with equations or something I may have missed. Please help me with the solving part!

 

[kuruman]

Yes, you need to solve a quadratic in $E_{\gamma}$ or $E_{photon}$ or whatever symbol you choose for the unknown photon energy. Use $\Delta E$ instead of 14.4 and put in the numbers at the very end.

 

[Rahulrj]

Yes, you need to solve a quadratic in $E_{\gamma}$ or $E_{photon}$ or whatever symbol you choose for the unknown photon energy. Use $\Delta E$ instead of 14.4 and put in the numbers at the very end.

I use $E_{photon}$ for the unknown. So the equation will be $2mc^2E_{photon}+E_{photon}^2 = \Delta E * 2mc^2$
The numbers at the right hand side is going to be too large mc^2 is in GeV and $\Delta E$ is in keV. Therefore on the left, 2mc^2 is written as .102*10^12 to be consistent with units.
$.102 * 10 ^{12} E_{photon}+E_{photon}^2 = 1477.4*10^{12}$ So I will get two values for $E_{photon}$ and they are
$E_{1}=14484.31$ and $E_{2} = -1.02*10^{11}$.
This looks strange to me probably because I haven't solved a quadratic with such large values and these values are clearly not among the options also it is to be noted that $E_{1}$ cannot be the energy as it is supposed to be lower than 14.4. I can't see where I went wrong. However, it is interesting to note that with the substitution of $E_\gamma$ (the known energy of photon) posted earlier I get an answer given among the options though the substitution itself doesn't make much sense to me. Still need help figuring this out.

 

[Rahulrj]

Yes, you need to solve a quadratic in $E_{\gamma}$ or $E_{photon}$ or whatever symbol you choose for the unknown photon energy. Use $\Delta E$ instead of 14.4 and put in the numbers at the very end.

I use $E_{photon}$ for the unknown. So the equation will be $2mc^2E_{photon}+E_{photon}^2 = \Delta E * 2mc^2$
The numbers at the right hand side is going to be too large mc^2 is in GeV and $\Delta E$ is in keV. Therefore on the left, 2mc^2 is written as .102*10^12 to be consistent with units.
$.102 * 10 ^{12} E_{photon}+E_{photon}^2 = 1477.4*10^{12}$ So I will get two values for $E_{photon}$ and they are
$E_{1}=14484.31$ and $E_{2} = -1.02*10^{11}$.
This looks strange to me probably because I haven't solved a quadratic with such large values and these values are clearly not among the options also it is to be noted that $E_{1}$ cannot be the energy as it is supposed to be lower than 14.4. I can't see where I went wrong. However, it is interesting to note that with the substitution of $E_\gamma$ (the known energy of photon) posted earlier I get an answer given among the options though the substitution itself doesn't make much sense to me. Still need help figuring this out.

 

[kuruman]

Both solutions that you propose are unacceptable. The photon energy cannot be greater that the difference between the excited and ground state energy ($E_1$) and it cannot be negative ($E_2$). It looks like you made a mistake somewhere which will be hard to detect with all those numbers floating about.
First solve the quadratic symbolically. Only one solution is acceptable, try to cast it in the form $E_{photon}=mc^2(something)$. Then worry about what is large and what is small and how to put in the numbers.

 

[Rahulrj]

I have not been able to solve this one yet. Since mass of photon is zero I don't get how casting it in the form you have mentioned will help.

 

[kuruman]

In the equation $E_{photon}=mc^2(something)$, $m$ is the mass of the atom. You derived this equation, so you ought to know what the symbols stand for.

 

[Rahulrj]

In the equation $E_{photon}=mc^2(something)$, $m$ is the mass of the atom. You derived this equation, so you ought to know what the symbols stand for.

$E_{photon}$ stands for the energy of the photon and therefore the mass will be associated to it isn't it ?! Maybe I may have misunderstood which equation you are referring to. I was referring to this equation $2m_ac^2E_{photon}+E_{photon}^2=\Delta E*2m_ac^2$ and took that you mentioned a substitution for $E_{photon}$ in terms of mc^2. In the equation I have shown $m_a$ stands for mass of the atom.

 

[kuruman]

In all your posts excluding #19, $m$ stands for the mass of the atom. The photon has zero mass. Please reread my post #16. I am asking you to solve a simple quadratic equation symbolically. Call the mass of the atom $m_a$ if you wish - I don't care. If you do call it $m_a$, then bring the acceptable solution of the quadratic into the form $E_{photon} = m_ac^2(something)$ where $(something)$ is an algebraic expression containing $\Delta E$, $m_a$ and $c^2$. It is the ratio of the photon energy to the rest mass energy of the atom.