How Does Radioactive Decay Affect Current in a Spherical Capacitor Circuit?

[HotFurnace]

Homework Statement

A Metallic sphere with radius r=10cm is surrounded by a spherical metallic shell of radius 2r=20cm. The space exist between the two is vacuum. An amount of k=0.01 mol of radioactive 242Cm is strayed uniformly on the sphere, the radioactive isotope emit alpha particles which have energy of E=6.1MeV upon decay. The decay half time of 242Cm is T=163days. Outer shell and inner sphere is connected to a resistor of resistance R, which then form a complete circuit. Assume that the resistor has no effect on the electric field between the sphere and the outer shell, and the metallic parts absorb all alpha particles that hits them. Consider the system at equilibrium, the current through the resistor is near constant. The charge of electron is e and Avogadro constant is Na
a/ Find the current through the resistor, when its value is R=100kΩ
b/ If we want the current to be one half the value in a/, then what's the resistance value?

Homework Equations

3. The Attempt at a Solution [/B]
We starts by finding the condition for an alpha particle emit at the surface of the inner sphere with angle α in respect with the surface normal vector, to reach the outer shell, apply energy conservation for that particle (assume the voltage between the sphere and the shell is U): $$2eU=\frac{m(vcos(α))^2} {2}=2eE*cos(α)^2$$
So all particles emitted in the angle of θ=2π-2arccos(α) will never reach the outer shell, in another word, they are absorbed by the inter sphere. Total charge emitted by the decay of the radioactive layer per unit time: $$\frac {dn} {dt}=-\frac {2keln(2)*Na*exp(-\frac {tln(2)} {T})} {T}$$ There's a 2 because the charge of alpha particle is 2e.
Note that the particles are emitted uniformly, so the charge collected by the inter sphere per unit of time is:
$$i=\frac {de} {dt}=-\frac {2keln(2)*Na*(2π-2arccos(sqrt(\frac {U} {E})))*exp(-\frac {tln(2)} {T})} {T}$$
This current is either discharged by the resistor, or charging the capacitor formed by the two metallic parts:
$$i=\frac {CdU} {dt} + \frac {U} {R}$$
Take note that the capacitance is $C=8πεr$
Solve the differential equation (I used initial condition U=0), we will obtain the result of: $$U=-\frac {2keln(2)*Na*(2π-2arccos(sqrt(\frac {U} {E})))*(exp(-\frac {tln(2)} {T})-exp(-\frac {t} {RC}))} {T*(\frac {1} {R}-\frac {Cln(2)} {T})}$$
We will still have to solve the result for U, and I can't do it. What shall I do?

 

[trurle]

You do not need the time-dependent solution - the question is about steady state circuit. The 100kOhm resistor for circuit is too small (you need about 65 GOhm for properly matching the direct charging generator), therefore the electric field in cavity is negligible, and nearly exactly 50% of alpha-particles (all outward-going ones) will be collected. My calculation for this case give 48uA current.

The half-current resistor is simply perfect power match resistor, ~65 GOhm at ~3.05MV voltage on it. You need to take spherical integral of radial component of energy of alpha-particles to get more accurate value for voltage at which 25% of particles still reach outer shell, to find precisely power-matching resistor value.

 

[HotFurnace]

Actually I want to solve generalized solution, so that I could take limits to get the answer. If you simplify $\frac {U} {E}≈0$ or $2arccos(sqrt(\frac {U} {E}))≈π$, the exponential component of $exp(-\frac {tln(2)} {T})≈1$ and $exp(-\frac {t} {RC})≈0$ for t in the range of minutes, then it will simplify to the answer of 47,5uA, this is of course the answer.
Thanks for the hint, I will try again and see how it go.

 

[TSny]

I stumbled on this patent https://patents.google.com/patent/US2598925A/en?oq=2,598,925 which you can download. I thought it was interesting even though I can't follow all the details.

 

[HotFurnace]

Sorry for reviving my old thread.If I use the hint, I will get 65GΩ as the answer, but the book states 105GΩ is the answer?? Something went wrong??

 

[TSny]

If I use the hint, I will get 65GΩ as the answer, but the book states 105GΩ is the answer?? Something went wrong??

I get an answer to (b) of 104 GΩ, which is close. Let α be the maximum angle from the radial direction that an alpha particle can leave the surface of the inner sphere and still make it to the outer shell. I found the value of α that would give a total current equal to half of 47.5 μA. I then found the potential difference between the inner sphere and the outer shell corresponding to this value of α.

The brown trajectory in the figure below corresponds to an alpha particle that leaves at the angle α, so it is just able to reach the outer shell. The blue trajectory doesn't make it. A key point is to note that for the brown trajectory, the kinetic energy of the alpha particle is not zero when it reaches the outer shell.