Question

Obtain the amount of Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of Co is 5.3 years.Refer image for question and equation.

Answer 1

given, strength of source , $\frac{dN}{dt}=8mCi$
we know, 1mCi = 3.7 × 10^7 d/sec
so, $\frac{dN}{dt}=8\times3.7\times10^7d/sec$
half life , t = 5.3 yrs
t = 5.3 × 365 × 24 × 60 × 60 sec

so, $\lambda=\frac{ln2}{T}=\frac{0.693}{T}$
= 0.693/(5.3 × 365 × 24 × 60 × 60)
= 4.14 × 10^-9

now, use radioactive formula,
$\frac{dN}{dt}=\lambda N$

8 × 3.7 × 10^7 d/sec = (4.14 × 10^-9)N

N = 7.15 × 10^16

so, mass of 7.15 × 10^16 atoms of Co.

= atomic mass × given number of atom/Avogadro's number
= 60 × 7.15 × 10^16/6.023 × 10^23
= 7.12 × 10^-6 g

Answer 2

Answer:

given, strength of source , \frac{dN}{dt}=8mCi

dt

dN

=8mCi

we know, 1mCi = 3.7 × 10^7 d/sec

so, \frac{dN}{dt}=8\times3.7\times10^7d/sec

dt

dN

=8×3.7×10

7

d/sec

half life , t = 5.3 yrs

t = 5.3 × 365 × 24 × 60 × 60 sec

so, \lambda=\frac{ln2}{T}=\frac{0.693}{T}λ=

T

ln2

=

T

0.693

= 0.693/(5.3 × 365 × 24 × 60 × 60)

= 4.14 × 10^-9

now, use radioactive formula,

\frac{dN}{dt}=\lambda N

dt

dN

=λN

8 × 3.7 × 10^7 d/sec = (4.14 × 10^-9)N

N = 7.15 × 10^16

so, mass of 7.15 × 10^16 atoms of Co.

= atomic mass × given number of atom/Avogadro's number

= 60 × 7.15 × 10^16/6.023 × 10^23

= 7.12 × 10^-6 g