How Does Beta Minus Decay Affect the Kinetic Energy of a Tin Nucleus?

[omiros]

Hello everybody, I am a first year student and I have a question that I am not sure if I can answer.

A radioactive isotope of Tin(50), with 85 neutrons in its nucleus and without any electrons bound to it, undergoes a beta minus decay. The electron pops into existence, 10^-15m from the center of the nucleus (consider that the nucleus is uniformly charged), and it leaves with a speed of 0.999c.

(We consider M_n=M_p=1.6x10^-27)

1. What is the final speed of, both the electron and the nucleus?

2. Will the the new nucleus, be able to bind the electron, into one of its orbits?

Tips: The new nucleus will have 51 protons and 84 neutrons.

The atom is a hydrogen like one, so we can find the probable bound energies through the Bohr model. (We don't care about its angular distribution)

The kinetic energy of the electron has to be calculated through the K = (γ-1)mc²
and the potential energy from the Coulomb law U = -Ze²/(4πε₀r).

I assume that the energy levels will be 51x13.6/n²eV.

So the hard part, is the equation of potential-kinetic energy with two varying speeds, that can give us the result.

 

[mfb]

"two varying speeds"?
Did you compare the kinetic energy of the electron to its potential energy?

(consider that the nucleus is uniformly charged)

That could indicate that you have to take the finite size of the nucleus into account. Do you know models about the size of nuclei?

 

[omiros]

"two varying speeds"?
Did you compare the kinetic energy of the electron to its potential energy?

No. I am just saying that, as the electron is travelling, it is 'pulling' the nucleus with it. A positive and a negative charge, that are getting separated (cause of the moving electron and the stationary, at the beginning nucleus.)

That could indicate that you have to take the finite size of the nucleus into account. Do you know models about the size of nuclei?

Personally I do not. However I am thinking of it as : It is of a finite size and quite small. 10^-15m, is its whole length.

 

[mfb]

No.

Well that would be interesting!

Personally I do not. However I am thinking of it as : It is of a finite size and quite small. 10^-15m, is its whole length.

In that case, the electron would "appear" (directly) outside the nucleus, and the given uniform charge distribution in the nucleus would not matter.

 

[paranormal]

First, let's calculate the initial kinetic energy of the electron. We can use the formula K = (γ-1)mc2, where γ is the Lorentz factor given by γ = 1/√(1-v2/c2). Since the electron's speed is given as 0.999c, we can calculate the Lorentz factor as γ = 1/√(1-0.9992) = 22.36. Plugging this into the formula, we get K = (22.36-1)(9.11x10^-31)(3x10^8)^2 = 2.46x10^-13 J.

Now, let's calculate the potential energy of the electron at a distance of 10^-15 m from the nucleus. We can use the Coulomb's law formula U = -Ze2/(4πε0r), where Z is the atomic number of the nucleus (in this case, 51), e is the elementary charge, ε0 is the permittivity of vacuum and r is the distance from the nucleus. Plugging in the values, we get U = -(51)(1.6x10^-19)^2/(4πx8.85x10^-12x10^-15) = -4.56x10^-12 J.

Now, we can use conservation of energy to calculate the final kinetic energy of the electron. The total energy (kinetic + potential) remains constant during the decay process. So, we can write the equation as Kinitial + Uinitial = Kfinal + Ufinal. Plugging in the values, we get Kfinal = Kinitial + Uinitial - Ufinal = 2.46x10^-13 + (-4.56x10^-12) - (-2.04x10^-13) = -4.72x10^-12 J.

Since the final kinetic energy is negative, we can conclude that the final velocity of the electron will be less than 0.999c. To calculate the exact velocity, we can use the formula K = (γ-1)mc2 and solve for v. This gives us v = √(1-(1-K/mc2)^2) = 0.998c.

Now, let's calculate the final speed of the nucleus. We can use conservation of momentum to write the equation as mvinitial = mvfinal. Since the mass of