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:
The kinetic energy of an alpha particle is the energy that it possesses due to its motion. It is typically measured in electron volts (eV) or mega-electron volts (MeV). For alpha particles emitted in radioactive decay, the kinetic energy usually ranges from about 4 to 9 MeV.
The kinetic energy (KE) of an alpha particle can be calculated using the equation KE = 1/2 mv^2, where 'm' is the mass of the alpha particle and 'v' is its velocity. For practical purposes in nuclear physics, the energy is often determined experimentally rather than through direct calculation.
The kinetic energy of an alpha particle is primarily affected by the type of radioactive decay process that produces it. The energy is determined by the difference in binding energy between the parent nucleus and the daughter nucleus plus the alpha particle. Other factors include the energy levels of the nuclei involved and the conservation of momentum and energy in the decay process.
The kinetic energy of an alpha particle is significant because it provides information about the energy released during nuclear decay processes. This energy can be used to understand the stability of nuclei, the forces at play within the nucleus, and the mechanisms of radioactive decay. Additionally, the kinetic energy of alpha particles is crucial in applications such as radiation therapy and nuclear energy.
Alpha particles generally have higher kinetic energy compared to beta particles but lower energy compared to gamma rays. However, because alpha particles are much more massive and carry a double positive charge, they have a much shorter range and higher ionizing power compared to beta particles and gamma rays. This makes alpha particles particularly effective in transferring energy to the material they pass through, causing significant ionization.