Question

A radioactive sample contains 2.2mg of pure¹¹C₆

which has half-life period of 1224 second. Calculate i) the no. of atoms present initially and ii) the activity when 5 microgram of the sample will be left.

Answer 1

Number of atoms resent would be

N = (avogadro's number X number of atoms) / weight

or

N = ( 6.023 X 10²³ X 2.2 X 10^-3 ) / 11

thus,

The number of atoms present initially

N = 1.2046 X 10²⁰

number of atoms remaining in 5 ug of the substance

N' = (avogadro's number X number of atoms) / weight

or

N' = N = ( 6.023 X 10²³ X 5 X 10^-6 ) / 11

thus

N' = 2.7377 X 10¹⁷

the radioactivity is given as

R = λN'

Where (for half life)

λ = 0.693 / T = 0.693 / 1224

or λ = 0.000566

thus,

R = 0.000566 X 2.7377 X 10¹⁷

so, the radioactivity would be, R = 1.55 X 10¹⁴ Bq