Questions
Determine the amount of necessary to provide a source of particles of 10 milli-curie strength if mean life of is
detailed solution
Correct option is C
Activity. A=dNdt=λN Given : A = 10 milli curie = 10×10-3×3.7×1010 dps λ=1Tm=11.72×107sec-1 we know that number of atoms =N=mass in grams xavagadro numbermolar mass No of atoms = N=Aλ=10×10-3×3.7×1010×1.72×107 ⇒N=6.36×1015 Mass of all these atoms = moles×molar mass = NNA×M ⇒Mass =2106×1023×6.36×1015=2.2×10−6gmTalk to our academic expert!
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