The discussion revolves around the kinetic energy of an alpha particle and its relationship with the thorium particle after a decay process. Participants explore concepts of conservation of energy and momentum in the context of nuclear physics.
The discussion is ongoing with multiple interpretations being explored. Some participants have provided guidance on writing separate equations for kinetic energy and using momentum conservation to relate the speeds of the particles. However, there is no explicit consensus on the correct approach or final answer yet.
Participants are navigating through the complexities of the problem, including the need to consider the initial state of the nucleus and the nature of energy transformations involved in the decay process. There are mentions of potential confusion regarding the definitions and implications of kinetic energy in this context.
T=1/2×4×Va²+1/2×mth×Vth²haruspex said:
You will need to tell me what all those symbols mean if you want me to check it.haha0p1 said:
Your approach is not correct because you did not use the fact that the momenta are equal in agnitude and opposite in direction. Furthermore, your conclusion does not follow from the work that you have shown. The last equation is an expression for the speed of the thorium. You have not explained your reasoning in terms of this equation that leads to your conclusionhaha0p1 said:
haha0p1 said:
Can you please explain to me how we got the right answer.?kuruman said:
Solving the question through your method:kuruman said:
You are still not using conservation of energy. What other energy change occurs in the reaction?haha0p1 said:
I am not getting your point. As far as I know, conservation of energy remains same at the end of the reaction. So T will equal the energy of alpha particle+energy of thorium. I am making the equation based upon this concept. Am I missing out something?haruspex said:
Where did that KE come from?haha0p1 said:
At first the nucleus was stationary. So, the energy T is in form of chemical energy. After the emission of Alpha particle and thorium, the energy is converting to mostly kinetic energy of the particle.haruspex said:
Not chemical.haha0p1 said:
The kinetic energy came from the internal energy of the nucleus?haruspex said:
Try nuclear.haha0p1 said:
Momentum of alpha particle=m×vharuspex said:
Try that one again.haha0p1 said:
I am unable to understand how we can use momentum equation to eliminate this one. Can you please explain.haruspex said:
Write the momentum equation correctly, then use it to express v1 in terms of the other variables. You can then use that to replace v1 in the energy expression.haha0p1 said:
That can't be right as the dimensions of the ##2v## are not right. The expression must have dimensions of energy.haha0p1 said:
You have reached the equation for the total energy that can be aritten as $$T=\frac{P_{Th}^2}{2m_{Th}}+\frac{P_{\alpha}^2}{2m_{\alpha}}.$$The problem is not asking you to find the kinetic energy of the alpha particle. It is giving you 4 choices. You can find the correct one by examining the equation above and using momentum conservation that says ##P_{Th}^2=P_{\alpha}^2.## If you cannot do this simply by looking at the equation, substitute ##m_{Th}=234~##u and ##m_{\alpha}=4~##u in the equation and figure out whether the second term is closer to ##T/2## than ##T##, i.e. if the answer one of (A, B) or one of (C, D). Then refine your choice.haha0p1 said:
To try to progress this, the conclusion is that ##v_1 \gg v_2##. I.e. if a small mass and a large mass have the same magnitude of momentum, then the small mass must have the greater speed.PeroK said: