# Obtain the amount of Co2760Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of Co2760Co is 5.3 years. - Physics

Numerical

Obtain the amount of ""_27^60"Co" necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ""_27^60"Co" is 5.3 years.

#### Solution

The strength of the radioactive source is given as:

"dN"/"dt" = 8.0 mCi

= 8 × 10−3 × 3.7 × 1010

= 29.6 × 10−7 decay/s

N = Required number of atoms

Half-life of ""_27^60"Co", "T"_(1/2) = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 108 s

For decay constant λ, we have the rate of decay as:

"dN"/"dt" = lambda"N"

Where, lambda = 0.693/"T"_(1/2) = 0.693/(1.67 xx 10^8) "s"^(-1)

∴ N = 1/lambda ("dN")/"dt"

= (29.6 xx 10^7)/(0.639/(1.67 xx 10^8))

= 7.133 × 1016 atom

For ""_27^60"Co"

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

∴ Mass of 7.133 × 1016 atom

= (60 xx 7.133 xx 10^16)/(6.023 xx 10^23)

= 7.106 × 10−6 g

Hence, the amount of For ""_27^60"Co" necessary for the purpose is 7.106 × 10−6 g.

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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.9 | Page 463
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 9 | Page 463

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