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Determine the amount of   84Po210  necessary to provide a source of α particles of 10 milli-curie strength if mean life of   84Po210 is 1.72×107 sec

a
2.2×10−7gm
b
2.2×10−9gm
c
2.2×10−6gm
d
2.2×10−10gm

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=2.2×10−6gm

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