# Gamma ray emission

by danielatha4
Tags: emission, gamma
 P: 113 1. The problem statement, all variables and given/known data A 67-Zn nucleus is at rest and in its first excited state, 93.3 keV above the ground state. The nucleus then decays to the ground state with the emission of a gamma ray. (One atomic mass unit is 931.5 MeV/c2.) What is the recoil speed of the nucleus? (You can assume non-relativistic motion, if needed.) 2. Relevant equations Kz=Pz2/2m Eg=Pgc Eg+Kz=93.3 keV Where Pz=Pg in magnitude 3. The attempt at a solution There's no point for me to type everything I did, but I'll describe it. I solved for P2 in each equation and set them equal to each other: Kz2m=Eg2/c2 then I used Eg+Kz=93.3 keV to substitute and I attempted to solve using the quadratic equation. However, a few times now I've gotten the Kz to be greater than 93.3keV by itself, which is obviously not right.
 P: 18 Bump. I'm having trouble with the same problem. It would be easy if it gave the energy of the photon, but it gives the change of energy between the nucleus and the photon. My line of thinking: 0=mv+E/c for momentum (delta)E=(1/2)mv^2 + pc for the energy I can't see where to go from here, and it might just be because it's 2:30, but any help is appreciated.
 P: 598 Hi, you need to use the formula for finding the recoil energy ER imparted on the nucleus. Think like this when a nucleus emits a gamma ray the nucleus recoil back..just find the recoil momentum, which is equal to momentum of gamma ray. for a starting point use this: $$E_{\rm R}=Mv_{\rm R}^2/2$$ and vR is recoil velocity (you need to solve for this quantity). Moreover when your nucleus emits gamma ray of energy 93.3 keV. and you also know how to find the recoil energy? if you know then you can solve for recoil velocity..got it? if you need some info. please reply.. I guess the solution is 448.663 m/s. good luck
P: 18

## Gamma ray emission

I understand that the recoil energy of the nucleus is described by (1/2)mv^2, and that the momentum of the photon and the nucleus are equal, I think my problem is not knowing the energy of the photon.
 P: 598 Hi Sarial, in your question i notice that the energy of gamma ray is 93.3 keV..[you have not read my previous post properly] So your velocity is 448.663 m/s.
 P: 18 I'm sorry. I did read your post properly. 93.3 keV isn't the energy of the photon though, it's the recoil energy of the nucleus and the gamma ray. Egamma + Knucleus = 93.3 keV Is that not right? Also, how do you get the 400 odd m/s? If 1/2Mv^2 = 93.3 keV 9.33e4 = .5(67(9.315e8))v^2 Solving for v would give 1.73e-3?
 P: 18 I tried something else that I think is along the right lines from what my teacher was saying, but he wasn't going too in depth at the time: My problem is 67.4 keV and a Nickel-61 nucleus Knucleus + Egamma = 67.4 keV pnucleus = -pgamma pnucleus + pgamma = 0 mv + E/c = 0 Knucleus = (1/2)mnucleusv^2nucleus (1/2)mnucleusv^2nucleus + Egamma = 67.4 keV 67.4 keV - (1/2)mnucleusv^2nucleus = Egamma So then plug the above equation for Egamma into the momentum equation mnucleusvnucleus + (67.4 keV - (1/2)mnucleusv^2nucleus)/c = 0 (931.5e6 eV x 61 amu)vnucleus + (67.4e3 eV/3e8 m/s)-((.5(931.5e6 eV x 61 amu)/(3e8 m/s))v^2nucleus Then solve for the quadratic: (5.68e10)vnucleus + 2.25e-4 - 94.7v^2nucleus (-94.7)v^2nucleus + (5.68e10)vnucleus + 2.25e-4 This gives you 6.00e8 m/s, which isn't right... Any ideas?
P: 598
 Quote by Sarial I'm sorry. I did read your post properly. 93.3 keV isn't the energy of the photon though, it's the recoil energy of the nucleus and the gamma ray. Egamma + Knucleus = 93.3 keV Is that not right? Also, how do you get the 400 odd m/s? If 1/2Mv^2 = 93.3 keV 9.33e4 = .5(67(9.315e8))v^2 Solving for v would give 1.73e-3?
Hi, absolutely you are wrong!
In the post it is mentioned clearly that 1st excited state is 93.3 keV. so why you get confusion here! that means 1st excited states has a energy of 93.3 keV and to go to ground state it needs 93.3 keV.....is the solutions correct or not ? the velocity is 448.663 m/s ?
 P: 598 Recoil energy $$E_{\rm R}=Mv_R^2/2$$. $$E_{\rm R}=E_{\gamma}^2/(2Mc^2)}$$. M is the mass fof zinc in kg and c is speed of light. $$E_{\gamma}$$ is 93.3 keV. solving you get that vel. is equal to 448.663 m/s
 P: 18 Awesome! That worked, thanks a lot. My one question is--where did the Erecoil = Egamma^2/(2Mc^2) come from? As in, what equation is it derived from?
 P: 598 Hi, when you apply conservation of momentum you easily derive that relation: $$E_{\rm R}=Mv_{\rm R}^2/(2M)=p_{\rm n}^2/(2M)$$ momemtum of $$\gamma$$-quantum is given by: $$p=E_{\gamma}/c=\hbar k$$ according to conservation of momentum we have $$|p_{\rm n}|=|p|$$. so $$E_{\rm R}=p_{\rm n}^2/(2M)=E_{\gamma}^2/(2Mc^2)}=(\hbar k)^2/(2M)$$ here $$p_{\rm n}, M$$ are momentum gained by the recoiling nucleus and mass of ISOTOPE in kg. You can use whatever you like. Tip: Apply with proper unit conversion (eV to J, etc ). Always remember recoil energies are generally very small compared to the energy of gamma quantum so you can treat the situation nonrelativistic approximation so all the above equations are valid. You can compute recoil energy (may be in meV) and compare with the energy of gamma quantum (in keV). Finally $$v_{\rm R}=E_{\gamma}/(Mc)=\hbar k/M.$$
 P: 18 Alright. Thanks a lot.
 P: 64 Thank you. I was stuck on a similar problem. Not sure why the Doppler effect gave me an extra factor of two; perhaps I should use the average velocity of the atom during the acceleration, and that would fix it...but I understand your derivation.
 P: 598 Hi Cruikshank, my post was for 'danielatha4'. Can you post your problem ? and just inform us where you struck. Then people here can help you.
P: 11
 Quote by Rajini Hi Sarial, in your question i notice that the energy of gamma ray is 93.3 keV..[you have not read my previous post properly] So your velocity is 448.663 m/s.
I know that this post is over a year old, but I saw this and felt I had to comment. The first excited state of the atom is 93.3 keV, as you said. When the atom decays into the ground state, however, this energy is not distributed completely to the gamma photon; some of the energy is imparted into the kinetic energy of the atom. It is clear that the atom must recoil to conserve momentum, and the recoil velocity must carry away some of the energy that was present in the excited state of the atom. If you based your result on the idea that the gamma ray has energy 93.3 keV, then your recoil velocity is incorrect.

Edit:
the gamma ray has 93.3 keV energy in the reference frame of the recoiling atom! Changing to the laboratory reference frame means that the gamma ray will lose energy due to the Doppler effect.
P: 11
 Quote by Sarial I tried something else that I think is along the right lines from what my teacher was saying, but he wasn't going too in depth at the time: My problem is 67.4 keV and a Nickel-61 nucleus Knucleus + Egamma = 67.4 keV pnucleus = -pgamma pnucleus + pgamma = 0 mv + E/c = 0 Knucleus = (1/2)mnucleusv^2nucleus (1/2)mnucleusv^2nucleus + Egamma = 67.4 keV 67.4 keV - (1/2)mnucleusv^2nucleus = Egamma So then plug the above equation for Egamma into the momentum equation mnucleusvnucleus + (67.4 keV - (1/2)mnucleusv^2nucleus)/c = 0 (931.5e6 eV x 61 amu)vnucleus + (67.4e3 eV/3e8 m/s)-((.5(931.5e6 eV x 61 amu)/(3e8 m/s))v^2nucleus Then solve for the quadratic: (5.68e10)vnucleus + 2.25e-4 - 94.7v^2nucleus (-94.7)v^2nucleus + (5.68e10)vnucleus + 2.25e-4 This gives you 6.00e8 m/s, which isn't right... Any ideas?
I can't go through algebra with the values already inserted... it's too messy... but here is my treatment on the relationship between the recoil energy and the excitation energy. I think I've shown that the former is definitely exceeded by the latter, using the quadratic equation approach you mentioned. Enjoy.

By conservation of momentum,

$P_{R} = P_{\gamma}$

$E_R = \dfrac{P_R^2}{2M}$

$E_\gamma = P_\gamma c$

$2ME_R = \dfrac{E^2_\gamma}{c^2}$

By conservation of energy,

$U = E_\gamma + E_R$

Here $U$ is the excitation energy.

$2ME_Rc^2 = (U - E_R)^2$

$2ME_Rc^2 = U^2 - 2UE_R + E_R^2$

$-\frac{1}{2}E_R^2 + E_R(Mc^2 + U) - \frac{1}{2}U^2 = 0$

Having solution

$E_R = (Mc^2 + U) \mp \sqrt{(Mc^2+U)^2 - U^2}$

$= U + Mc^2 \mp \sqrt{(Mc^2)^2 + 2Mc^2U}$

Now, let's express all energy as a proportion of $U$.

$\dfrac{E_R}{U} = 1 +\dfrac{Mc^2}{U} \mp \sqrt{\left [ \dfrac{(Mc^2)}{U} \right ]^2 +2 \dfrac{ Mc^2}{U} }$

We know that $Mc^2 \gg U$, so set $x \equiv \dfrac{Mc^2}{U}$

$\dfrac{E_R}{U} = 1 + \left [ x - \sqrt{x^2+2x} \right ]$

For x very large, this is

$= 1 + \left [ x - (1+\varepsilon) x\right ]$

where $0<\varepsilon < \frac{1}{x}$. Thus,

$\dfrac{E_R}{U} = 1 - \varepsilon x < 1$

so the recoil energy is less than the excitation energy.

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