Recent content by moenste

Beta decay is better for unstable nuclei + plus alpha-decay is usually used for "heavy" nuclei. 60Co → 60Ni + 0-1β. I think this should be correct.

OK, now I understand what you mean. But if we have Co-60 that decays, we can use any kind of decay (alpha, beta, gamma, etc.). What specific kind do they suggest?

But it says that. We have Co-60 that decays to Co-59. Or I understand the text wrong?

mD vD = ma va KED + KEa = 5.404 MeV KED = 1/2 mD vD2 vD = ma va / mD KED = 1/2 mD ma2 va2 / mD2 KEa = 1/2 ma va2 KED = 1/2 ma va2 * ma / mD KED = KEa * ma / mD KEa * ma / mD + KEa = 5.404 MeV Take KEa out of the brackets: KEa (ma / mD + 1) = 5.404 MeV KEa = 5.404 MeV / (ma / mD + 1) I can't...

Yes, the (c) part is clear. I used p = n R T / V. Mostly was confused on the theory part with atoms.

p = m v = 0 p = m valpha + m vnucleus = 206 vnucleus + 4 valpha 0 = 206 vnucleus + 4 valpha 206 vnucleus = - 4 valpha I think this is unnecessary complication. The momentum, the velocities. More chances to do a mistake somewhere. We have Co-60 that becomes Co-59. I did that and used neutron...

I think I get it. The thing with Avogadro is not relevant for this problem, right? We don't want to know the number of atoms in 4 g of He-4, but we need to know the total number of atoms. One alpha particle makes one He-4 atom. But if we need to know the number of atoms in 4 g of He-4 atom...

Oh, yes, I looked at C but somehow wrote I.

Yes, completely forgot about the integer values. But how to derive the momentum then (velocity is unknown)? Plus in KE = p2 / 2 m I think m only includes one mass of either the alpha particle or the produced nucleus. As I understand the original isotope Co-60 has one more neutron than Co-59...

Yes, that's an improvement. "I" could be skipped and connected directly to "F". Thank you for the tip : ).

Time 0: p = m v = m * 0 = 0. Time 1: pα = mα vα, pnucleus = mnucleus vnucleus. They both have some kinetic energy and that is why the energy in (i) does not all appear as kinetic energy Eα? Total energy released = 5.404455 MeV. Total momentum = 0. Particle masses = 205.929 421 u and 4.001 594...

Thank you for the link. I wish I saw it earlier, could save me much time yesterday : ). The table has 6 columns between C and J, I don't think I would manage to get the answer by going backwards. Again, thank you for the link!

I understand the concept of momentum conservation. So p = m v initial and then find the velocities of the alpha particle and the nucleus?

But if one alpha particle is one He-4 atom so each 4 g of He-4 atoms have NA [6 * 1023 number] of atoms? So one atom has 6 * 1023 atoms? To me it looks like "a chair of a chair is a chair" o_O.

I have no idea. I searched for "kinetic energy alpha particle" and got tp this topic, but I don't have the mass and velocity. Found pictures like this and tried to calculate the answer using 209.936 730 - 205 - 4 and that didn't help. Integer values says that the number should have no digits...