Nuclear Physics. Help Me Solve This Question.?

The effective half-life is the reciprocal of the sum of the reciprocals of the biological and physical half-lives.

1/Teff = 1/Tphy + 1/Tbio

= 1/8 + 1/3 = 0.4583

Teff = 1/0.4583 = 2.18 days.

11 days = 11 / 2.18 = 5.04 times the effective half-life. Expect to see just under 1/32 of the initial activity after 11 days.

In fact, 2^(-5.04) = 0.0304

100mCi * 0.0304 = 3.04 mCi

i provides trend of reactor for plutonium. Output propose all potential that devour in it ( in rays). There U235 emit the neutrons(photons) that devour by using U238 to get Pu239. First there is only U238. For reaction is going want U235 that prods while U238 getting its critics mass and serve afterwords as gasoline. via very organic components themperature in this reactors 70 degree (chilly sintez) with 50MW potential. This potential count selection U238 and Pu it relatively is virtually a similar mass. in one day that's one hundred gramm for 50MW. i don't be conscious of frequence of organic U238 emission to tell quantity of atoms and photons in accordance to Plank's constanta. i could greater advantageous culculate the potential in accordance to formulaes yet to not temperature of wapor and output electrical energy. So 100g*14= a million.4kg +0.3kg.=a million.7kg. i'm able to mistake with potential yet way is nice.