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What is a balanced nuclear equation for the alpha decay of Rn-22?
I'm assuming we are talking about Rn222. Rn222 is part of the U238 chain (also known as the Radium series). Rn222 (half life 3.8 days) -> Po218 + alpha particle (5.59 MeV) Po218 (half life 3.1 minutes) -> Pb214 + alpha particle (6.115 MeV) Pb214 (half life 26.8 minutes) -> Bi214 + beta- particle (1.024 MeV) Bi214 (half life 19.9 minutes) -> Po214 + beta- particle (3.272 MeV) Po214 (half life .164 ms) -> Pb210 + alpha particle (7.883 MeV) Pb210 (half life 22.3 years) -> Bi210 + beta- particle (64 keV) Bi210 (half life 5.013 days) -> Po210 + beta- particle (1.426 MeV) Po210 (half life 138.4 days) -> Pb206 + alpha particle (5.407 MeV) Pb206 (lead) is stable so that is the end of the chain.
What is a neutron initiator?
hi it is a combination of Be/Po210 . po210 emits an intensive alfa radiation and when this radiation contact with Be metal foil, the result would be neutron. as you know neutron is critical element to make chain fission reaction .so in practice a very thin gold foil can stop alfa radiation but if an explosion mix po210 with Be, we will have a high intense neutron source can trigger a ultimate fission bomb. a good design of initiator help to make a smaller and effective bomb with high yeild. the size of an initiator would be a grape size. if you have any more question don't hesitate to ask contact mail:annafarahmand@yahoo.com
What is the nuclear equation for the decay of Po210 if it undergoes 2 alpha decay then a beta decay followed by another alpha decay.?
The nuclear equation for the decay of Po-210 undergoing 2 alpha decays followed by a beta decay and another alpha decay is: Po-210 -> Pb-206 + 4 He-4 + 2 e-1 + 2 v This equation represents the series of decays that result in the transformation of Po-210 into Pb-206, with the emission of two helium nuclei (alpha particles), two electrons, and two neutrinos.
