What is the amount of \(^{60}_{27}Co\) necessary to provide a radioac

1.	What is the amount of \(^{60}_{27}Co\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Answer» Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci = (10.0 × 10^-3 )(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸dis/s T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷ ) s = 1.673 × 10⁸ s Decay constant, \(\lambda\) = \(\cfrac{0.693}{T_{1/2}}\) = \(\cfrac{0.693}{1.673\times10^8}s-1\) =4.142 × 10^-9 s^-1 Activity = Nλ ∴ N = \(\cfrac{activity}{\lambda}\) = \(\cfrac{3.7\times10^8}{4.142\times10^{-9}}\) atoms = 8.933 × 10¹⁶atoms = 60 grams of \(^{60}_{27}Co\) contain 6.02 × 10²³ atoms Mass of 8.933 × 10¹⁶atoms of \(^{60}_{27}Co\) = \(\cfrac{8.933\times10^{16}}{6.02\times10^{23}}\) x 60 g = 8.903 × 10^-6 g = 8.903 µg This is the required amount.

What is the amount of \(^{60}_{27}Co\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?

Answer»

Data : Activity = 10.0 mCi = 10.0 × 10^-3 Ci

= (10.0 × 10^-3 )(3.7 × 10¹⁰) dis/s = 3.7 × 10⁸dis/s

T_1/2 = 5.3 years = (5.3)(3.156 × 10⁷ ) s

= 1.673 × 10⁸ s

Decay constant, \(\lambda\) = \(\cfrac{0.693}{T_{1/2}}\) = \(\cfrac{0.693}{1.673\times10^8}s-1\)

=4.142 × 10^-9 s^-1

Activity = Nλ

∴ N = \(\cfrac{activity}{\lambda}\) = \(\cfrac{3.7\times10^8}{4.142\times10^{-9}}\) atoms

= 8.933 × 10¹⁶atoms

= 60 grams of \(^{60}_{27}Co\) contain 6.02 × 10²³ atoms

Mass of 8.933 × 10¹⁶atoms of \(^{60}_{27}Co\)

= \(\cfrac{8.933\times10^{16}}{6.02\times10^{23}}\) x 60 g

= 8.903 × 10^-6 g = 8.903 µg

This is the required amount.