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Gravitational redshift inside a crystal - Pound-Rebeka experiment

  1. Jul 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Some atoms emit X-rays at transition from excited to ground state. Energy of an excited state of the atom ##\scriptsize^{57}_{26}\textrm{Fe}## is ##\scriptsize14.4keV## larger than than the energy of it's ground state. The mass of the atom is ##\scriptsize9.5\times10^{-26}kg##. How much energy that is released during the transition from the excited to the ground state is carried away by the core? What is the ##\scriptsize\Delta \lambda## of the emitted photons? Some crystals are bound so hard that the momentum during transition is carried away by the whole crystal. How many energy is carried away by the crystal of mass ##\scriptsize 1g##? During the experiment which confirmed the existence of the gravitational redshift they used same crystal as a source of monochromatic light while they set same crystal ##\scriptsize 20m## away to absorb the emitted photons. What would be the ##\scriptsize\lambda## of emitted photons from the first crystal and how much would it change because of the change of gravitational energy? Was the width of the emission line wide enough in order to measure the gravitational
    redshift?

    2. Relevant equations
    I have been solving a problem set for the "photon in the gravitational field" and came across this problem. It looks very hard to solve and so far the only equation I needed to solve these types of problems was:

    \begin{align}
    \lambda_2 = \lambda_1 \left(1-\frac{GM}{c^2r}\right)
    \end{align}

    3. The attempt at a solution

    I presume that when an atom goes from excited to ground state it would emmitt light of ##\lambda##:

    \begin{align}
    \Delta W = h\nu = \frac{hc}{\lambda} \longrightarrow \lambda = \frac{h c}{\Delta W} = \frac{6.626\times10^{-34}Js \cdot 2.99\times10^8 \tfrac{m}{s}}{14.4\times10^{3} \cdot 1.602\times10^{-19}J} = 85.9pm
    \end{align}

    But in our case ##\lambda## will be a bit longer because i know that some of the energy ##\scriptsize\Delta W## will be caried away by the core - through gravitational potential energy i presume (can i use a whole mass of the atom here?). I could calculate the right ##\lambda## if i only knew how much does a radius ##r## change. How do i get radius?
     
  2. jcsd
  3. Jul 21, 2013 #2

    mfb

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    For the emission process itself, gravity is not relevant. The photon has a momentum, so the atom (or the whole crystal in the second case) has to carry the opposite momentum. Use energy/momentum conservation to find the energy of the photon.

    Which radius??
     
  4. Jul 21, 2013 #3
    Is that because the mass of the atom is too small?

    Thank you. I will try this and report back.

    I was thinking of a planetary model of an atom where the atom is circling around the core. I thought if i could calculate the change of the radius because of the energy change i could calculate the gravitational redshift. But as you said gravitational redshift can be neglected in the atom.
     
  5. Jul 21, 2013 #4

    mfb

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    No, you just don't have any relevant height differences within an atom.

    Eh... no.
     
  6. Jul 21, 2013 #5
    Ok so at start i have only the rest energy of the ##\scriptsize ^{57}_{26}Fe## atom while after colission i am not sure what i have. In other words i do not know how to write down that the rest energy is lower. Can i do it like this?

    \begin{align}
    E_{before} &= E_{after}\\
    E_{0~Fe} &= \sqrt{\smash{\underbrace{(E_{0~Fe} - E_\gamma)^2}_{\substack{\text{I think of this}\\\text{as the rest energy}\\\text{lowered because of}\\\text{the emission of a photon}}}} + p^2c^2}\longleftarrow\substack{\text{is momentum here equal to $p \, =\, (E_{0~Fe}-E_\gamma)\,/\,c$ ???}}
    \end{align}
     
    Last edited: Jul 21, 2013
  7. Jul 21, 2013 #6

    mfb

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    After the decay, you have the rest energy of the iron, the photon energy and the kinetic energy of the iron atom.

    The rest energy of the iron, compared to its initial energy, went down by 14.4 keV. Those 14.4 keV got converted to the photon energy and the kinetic energy of the iron atom.
     
  8. Jul 21, 2013 #7
    So i presume my equation is correct if i set ##p = \frac{E_{\gamma}}{c} = \frac{14.4keV}{c} ## ?
     
  9. Jul 21, 2013 #8
    Okay so the emitted energy got converted into an energy of a photon and kinetic energy - this means energy of the photon ##E_\gamma < 14.4keV##. I tried to calculate it like this and got nonsense:

    \begin{align}
    E_{1} &= E_{2}\\
    \sqrt{ {E_{0~Fe}}^2 + {p_1}^2c^2 } &=\sqrt{ \left(E_{0~Fe} - E_\gamma\right)^2 + {p_2}^2 c^2 } \longleftarrow \substack{\scriptsize \boxed{p_1 = 0}~\boxed{p_2 = E_\gamma /c}}\\
    \sqrt{ {E_{0~Fe}}^2 + 0 } &= \sqrt{ \left(E_{0~Fe} - E_\gamma\right)^2 + \frac{{E_\gamma}^2}{c^2} c^2 } \\
    E_{0~Fe} &= \sqrt{ \left(E_{0~Fe} - E_\gamma \right)^2 + {W_\gamma}^2 }\\
    {E_{0~Fe}}^2 &= {E_{0~Fe}}^2 - 2E_{0~Fe}E_\gamma +{E_\gamma}^2 + {E_\gamma}^2 \\
    {E_\gamma}^2 + (-E_{0~Fe})E_\gamma + 0 &= 0\\
    &\Downarrow\\
    E_\gamma &= 0\\
    E_\gamma &= E_{0~Fe}
    \end{align}

    None of the solutions to the quadratic equation make sense. 1st solution is trivial and 2nd is way too big.
     
    Last edited: Jul 21, 2013
  10. Jul 21, 2013 #9

    mfb

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    Your photon is missing in this equation.
    You don't have to use relativistic equations here - the iron atom will be slow. In addition, you don't have to consider the initial situation.

    As function of the velocity and the mass, what are energy and momentum of the iron atom (after the emission)?
    What is the relation between energy and momentum for the photon?

    The sum of the two momenta has to be zero, and the sum of the energies has to be 14.4 keV. Solve for the photon energy, and you are done.
     
  11. Jul 21, 2013 #10

    haruspex

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    Please define your variables. E.g. what is p1? Why can't I see any reference to the momentum and velocity of the atom after emission?
     
  12. Jul 21, 2013 #11
    I did it relativisticaly like this:

    \begin{align}
    E_1&=E_2\\
    E_{1~atom}&=E_{2~atom} + E_\gamma\\
    E_{0~Fe} + \Delta E &= \sqrt{{E_{0~Fe}}^2 + {p_2}^2c^2} + E_\gamma\\
    E_{0~Fe} + \Delta E &= \sqrt{{E_{0~Fe}}^2 + \frac{{E_\gamma}^2}{c^2}c^2} + E_\gamma\\
    E_{0~Fe} + \Delta E - E_\gamma &= \sqrt{{E_{0~Fe}}^2 + {E_\gamma}^2}\\
    \substack{\text{square of}\\\text{a trinomial}}\longrightarrow(E_{0~Fe} + \Delta E - E_\gamma)^2 &= {E_{0~Fe}}^2 + {E_\gamma}^2\\
    \big((E_{0~Fe} + \Delta E) - E_\gamma\big)^2 &= {E_{0~Fe}}^2 + {E_\gamma}^2\\
    (E_{0~Fe} + \Delta E)^2 - 2(E_{0~Fe} + \Delta E)E_\gamma + {E_\gamma}^2 &= {E_{0~Fe}}^2 + {E_\gamma}^2\\
    (E_{0~Fe} + \Delta E)^2 - 2(E_{0~Fe} + \Delta E)E_\gamma &= {E_{0~Fe}}^2\\
    E_\gamma &= \frac{(E_{0~Fe} + \Delta E)^2 - {E_{0~Fe}}^2}{2(E_{0~Fe} + \Delta E)}\\
    E_\gamma &= \frac{\left(53\times10^9eV + 14.4\times10^3eV\right)^2 - \left(53\times10^9eV\right)^2}{2\left(53\times10^9eV + 14.4\times10^3eV\right)}\\
    E_\gamma &= 14.399 keV
    \end{align}

    The result is ok now thanks to your comments, but it surprised me the atom takes away so little energy!
     
  13. Jul 21, 2013 #12

    mfb

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    The atom is heavy, so it can get a lot of momentum with a tiny amount of energy.
     
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