Calculations in nuclear physics

Hello!

I need help in calculating certain values in nuclear physics. Here are the tasks:

1.) For the radioactive decay of the element polonium, it shall be deems

$^{210}_{\ 84}\mathrm{Po} \to{}^{206}_{\ 82}\mathrm{Pb} + {}^{4}_{2}\mathrm{He}$

The half period of Polonium is T = 138.4 days. The mass of the polonium at the beginning of the decay is M = 0.3 mg. The relative masses m of polonium and helium are Po: 210 and He: 4,003

What is the mass M of helium after two years?

Before I calculate the mass M of helium, I have to calculate the mass M of polonium after two years first. I thought it would be possible to determine the mass M of poloium by the law of decay:

where $\lambda = \dfrac {ln 2}{T}$, t = 780 days (= two years) and $N_{0}$ = 0.3 mg.

And after calculating the mass M of polonium, the mass M of helium can be calculated by:

$\dfrac{M(Po)}{m(Po)} = \dfrac{M(He)}{m(He)}$

$M(He) = \dfrac{M(Po)*m(He)}{m(Po)}$

Is this solution process right?

2.) A radioactive preparation has the half period of 10 minutes. Determine the decay konstant $\lambda$, that is to say the fraction of the decaying atoms per second in percent (the percent rate of decaying atoms in a second).

I guess it has to be:

$\lambda = \dfrac {ln2}{T}$, where T = 10 minutes = 600 seconds.

Multiplied by 100, I get the percent rate. Is that right?

1. what do the upper and lower case M's represent here? (...)

(...)
fwiw I think they've provided the mass of helium and the mass of polonium because they don't want you to say "mass decrease of polonium" = "mass increase by helium",I think they want you to calculate the number of Po nuclei decaying over 2 years and multiply that by the mass of helium. (...)

Oh right - have you subtracted the mass of polonium at t=2yrs from the mass of polonium at t=0 ?

No, I have not. Do you mean I should subtract the mass of polonium at the beginning from the mass of polonium after two years? how can I calculate the mass of polonium after two years?

And after I have done it, the mass of helium can be calculated by the equation above, so by the masses and relative masses of polonium and helium?

x

If I put the values above in the law of decay, I get $N = 0,061$ mg (rounded) as decaying mass of polonium after two years. Substract from the mass at the beginning, I get:

0,3 mg - 0,061 mg = 0,239 mg (rounded) as mass of polonium after two years. Is this calculation right?

I don't get that - can you show your working? (...)

As you will!

I have calculated the mass after the law of decay: , where:

$\lambda = \dfrac{ln2}{T}$

$T = 138,4$

$t = 730$ (= 2 years)

$N_{0} = 0.3$mg

Put in the law of decay, I get:

N = 0,061 mg (rounded)

Subtracted to the mass of polonium at the beginning of the decay, I get: 0,3 mg - 0,061 mg = 0,239 mg

Why is this result wrong? Do I have this right that the mass of helium can be calculatred by the relative masses and the mass by that:

$M(He) = \dfrac{M(Po)*m(He)}{m(Po)}$

If that is the case, the mass of Helium is:

$M(He) = \dfrac{0,239*4,003}{210}$ = 0,0046 mg (rounded)

(...)

QUICK CHECK

730/138.4=5.27 (number of half periods)

so we'd expect the amount in mg to be somewhere between
0.3/(2^5) = 0.00937 and 0.3/(2^6) = 0.00468

(...)
0.3x0.02584
=0.007751 (mg)

That makes sense, as there are 5 up to 6 half periods in two years.





Why do you multiply the mass 0.3 mg with 0.2584 mg instead of to substract?

If I put your calculated value in the equation $M(He) = \dfrac{M(Po)*m(He)}{m(Po)}$ I get:

$M(He) = \dfrac{0.007751*4.003}{210}$ = 0.00015 mg (rounded)

Your result is not between 0.00937 and 0.00468, in contrast to mine, as I got 0.0046. So I suppose that your calculation must be wrong.

(...)
I multiply 0.3mg by 0.02584 (not 0.2584) because e to the power of -λt is 0.02548 and the decay law is telling us to multiply the original quantity by e to the power of -λt to get the quantity remaining at time t