Q - What is the wavelength of the incoming photon?

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The discussion centers on calculating the wavelength of an incoming photon that strikes a helium nucleus, resulting in the formation 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). The relativistic factor, gamma (γ), is utilized to account for the energies involved, specifically in the context of the helium nucleus being initially at rest. The participants confirm the necessity of using rest energy for the helium nucleus in the calculations.

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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/[tex]\lambda[/tex] = 2([tex]\gamma[/tex]m(H)c^2) - ([tex]\gamma[/tex]m(He)c^2)

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

Then I would solve for [tex]\lambda[/tex] 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 [tex]\gamma[/tex] in the helium associated term. And even if it wasn't at rest, it wouldn't have the same [tex]\gamma[/tex] 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, [tex]\gamma[/tex] would not be the same. Sorry about that!
 

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