B Decay of Excited Atoms: Probability & Nonlinearity

Trollfaz
Messages
143
Reaction score
14
I learned that the probability of radioactive decay for an atom is always the same. However, is the decay of and excited atom or electron non linear(decay probability varies with time)?
 
Physics news on Phys.org
Radioactive decay is governed by the exponential law $$N(t)=N_0e^{-\lambda t}$$ where N_0 denotes the number of nuclei at t=0 , and \lambda is the decay constant which is different for different nuclei. If you use the word "probability of radioactive decay" for "rate of radioactive decay" it is not same for all nuclei. In case of radioactive decay too, the decay rate \frac{dN}{dt} also varies with time in a nonlinear fashion.

In presence of the environment, the eigenstates of the atomic Hamiltonian are not true eigenstates of nature. Therefore, each atomic level has a certain lifetime \tau which depends upon the interaction it is subjected to. In general, a single atom continues to make transitions between the atomic energy levels.
 
Last edited:
I mean is decay of excited atoms or electrons governed by the exponential law too?
 
The probability that an existing atom will decay within the next second is always the same.

If you follow a speific atom, the probability that it exists (in the original state) goes down over time - because it might have decayed. That means the probability that you observe the decay after X seconds is smaller than the probability that you observe it after Y seconds if X>Y.Every radioactive decay follows an exponential decay.

There are no excited electrons. Excited atoms exist, and if you don't influence them from the outside, they will decay exponentially.
 
  • Like
Likes stoomart
Trollfaz said:
I mean is decay of excited atoms or electrons governed by the exponential law too?

If at any instant of time all the atoms are excited to the higher energy level i.e., a population inversion is achieved, it will eventually relax back to a Boltzmann distribution in presence of a temperature bath.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top