Q - What is the wavelength of the incoming photon?

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The discussion revolves around calculating the wavelength of an incoming photon that strikes a helium nucleus, resulting in the creation of two hydrogen isotopes moving at 0.6c. The conservation of energy principle is applied, leading to the equation E(photon) = 2E(hydrogen) - E(helium). Participants clarify that the helium nucleus is initially at rest, which simplifies the energy calculations to its rest energy E = mc^2. There is a focus on ensuring the correct application of the relativistic gamma factor, as it differs for the hydrogen isotopes compared to the helium nucleus. The conversation emphasizes the importance of correctly identifying the initial conditions of the helium nucleus for accurate problem-solving.
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An energetic photon strikes a helium nucleus and breaks into two hydrogen isotopes (A = 2, Z = 1). The two particles move off at 0.6c. Assume all energy of the photon is absorbed and that relativistic energies are involved.

Q - what is the wavelength of the incoming photon?

Conservation of energy - Ei = Ef
E(photon) + E(helium) = 2E(hydrogen)

E(photon) = 2E(hydrogen) - E(helium)

hc/\lambda = 2(\gammam(H)c^2) - (\gammam(He)c^2)

\gamma = 1/sqrt(1-u^2/c^2)

Then I would solve for \lambda by rearranging the equation.
Is this how i should be solving this type of problem?

Thanks
DoubleMint
 
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Hello DoubleMint,

Was the Helium nucleus (which exists before the collision) initially at rest? You have a \gamma in the helium associated term. And even if it wasn't at rest, it wouldn't have the same \gamma as the one associated with the hydrogen nuclei, would it?
 
Hi collinsmark,

The helium nucleus is initially at rest, so would it be just be the rest energy of the particle, E = mc^2?
If it was not at rest, \gamma would not be the same. Sorry about that!
 
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