Find the speed of a recoiling atom(modern physics)

[Raziel2701]

Homework Statement

An excited iron nucleus (Fe) of mass 57u decays to its ground state with the emission of a photon. The energy available from this transition is 14.4 keV. By how much is the photon energy reduced from the full 14.4 keV as a result of having to share energy with the recoiling atom?

The Attempt at a Solution

Everyone tells me to use conservation of momentum but I don't know how to set this up. Initially the iron nucleus is at rest so the momentum is zero. Then the emission of the photon occurs and so I should have the momentum of the photon plus the momentum of the nucleus equal zero right?

But the photon's momentum if Planck's constant divided by the wavelenght, which I don't know, so I have two unknowns and one equation, I can't solve that.

So what do I do?

 

[vela]

Conservation of energy gives you a second equation.

 

[Raziel2701]

So on one side of the equation I would have 14.4 keV= the kinetic of the Fe nucleus plus the energy of the photon. I see now three unknowns, the velocity of the nucleus after photon emission, the wavelength of the photon and the energy of the photon.

 

[sgd37]

I would think the magnitude of momentum is equal to that of the photon. Therefore the atom would have the entire energy of the system if it had double its momenta

 

[vela]

So on one side of the equation I would have 14.4 keV= the kinetic of the Fe nucleus plus the energy of the photon. I see now three unknowns, the velocity of the nucleus after photon emission, the wavelength of the photon and the energy of the photon.

The energy and momentum of a photon are related by E=pc.

 

[Raziel2701]

So I've done the following:

Conservation of momentum
0=(57u)v - \frac{h}{\lambda} to solve for the velocity of the iron nucleus of mass 57u. I then plugged it into the equation for conservation of momentum, essentially 14.4keV equals the kinetic of the nucleus and the energy of the photon(hc/lambda):

14.4keV=.5(57u)(\frac{h}{\lambda 57u})^2 +\frac{hc}{\lambda}

So I was able to solve for lambda, the wavelength, and I got .86277 angstrom but then I solved for velocity and I got a velocity 3 orders of magnitude greater than the speed of light...

So obviously I'm doing something wrong, I keep reading the comments that people have left but I'm not getting them. I don't know how to set this problem up.

 

[vela]

How are you calculating the velocity?

 

[Raziel2701]

With the first equation on my last post, conservation of momentum I have that 0= (57u)v - h/lambda. Having found lambda I just plug it in and solve for v.

 

[vela]

I suspect you're just screwing up the units, but it's hard to say unless you show us the actual calculations.

 

[Raziel2701]

So in your opinion my steps, my set up I've placed here is correct? I'll solve it again, see what comes up.

 

[vela]

Yeah, it looks fine to me. I got the same photon wavelength and found a speed of about 81 m/s when I worked it out.

 

[Raziel2701]

Yeah I know what I did wrong. I was solving the energy equation symbolically through matlab, and I had some errors here and there. I got the right answer like you did, at first I thought it didn't make sense to have such velocity(seemed too large to me) but once I calculated what fraction of the total energy (14.4 keV) it represented, I felt at ease again.

Thanks!