I am trying to resolve a question about Relativity Theory, with an answer key at hand. The answer given creates more confusion than it resolves, and I am looking for some insights and answers to very specific questions.
Radioactive nucleus A* is at rest within the framework of a laboratory. It collapses into nucleus A with rest mass $m_o$; and into electromagnetic radiation $γ$ with energy $E_γ$ within the framework of the laboratory.
Use $E_γ$ and $m_o$ to express the following:
- the rest mass of $A*$
- the velocity of $A$ within the framework of the laboratory
- the frequency of the photon within A's framework.
I have various issues with the answer-key, and by posting this here hope to understand what I am misinterpreting:
- According to the Law of Conservation of Energy: $E_γ+m_0 c^2=E_(A*)$
My problem here is that I do not understand why A's energy is described as $m_0 c^2$. The electromagnetic radiation has no rest-mass, so its energy is equal to $P·c$, which implies that $A$ should have the opposite momentum, given that A* was at rest (i.e. momentum = o). Indeed, nowhere was there mention of $A$ being at rest, so why should I suppose it was?
- $p = \frac{E}{c} ⇒ mvγ=\frac{E}{c} ⇒ v=\frac{(m_0 c^2)}{mγc}=\frac{c}{γ}$
I assume that $p$ here refers to A, given the nature of the question. Yet, this clashes with the previous answer, in that suddenly A does seem to have momentum (indeed it explicitly has velocity now), and this time $m_0 c^2$ is nowhere to be found in the definition of energy.
- $E=hν=m_0 c^2⇒ν=\frac{(m_0 c^2)}{h}$
Here I am at a loss how and why $hν=m_0 c^2$, nor do I see how this relates to A's framework.
Many thanks in advance!
