Question

Determine the mass of Na22 which has an activity of 5mCi. Half life is 2.6 years

Answer 1

Answer:

Mass of Na22 is $8.4 \times 10^{-6} g$

Explanation:

We know that  $1 Ci=3.7\times 10^{10} dps$

Activity of Na22 is 5m Ci is A=$3.7\times 10^{10} \times 5 \times 10^{-3}=1.85 \times 10^8 dps$

$Half life =2.6 \times 365 \times 24 \times 60 \times 60=8.2 \times 10^7 s$

From radioactive decay equation, activity of radioactive substance $A = \dfrac{0.693}{T_{1/2}} N$

$1.85 \times 10^8 =\frac{0.693}{8.2 \times 10^7} N\\\\N= 2.2 \times 10^{16}$

Number of Na22 atoms is $2.2 \times 10^{16} Na22 atoms$

Atomic weight of Na22 is 23g

mass of Na22 is

$m=\frac{N}{N_{A}} M\\m=\frac{2.2 \times 10^{16}}{6.023\times 10^23}23\\\\m=8.4 \times 10^{-6} g$

Mass of Na22 is $8.4 \times 10^{-6} g$