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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?