Question

Show that the mass of radium \[\ce{^226_88Ra}\] with an activity of 1 curie is almost a gram. Given T1/2 = 1600 years.

Answer 1

Activity = 1 curie

Half life of Radium = 1600 years

1 curie = 3.7 × 10¹⁰ disintegrations per second

From law of disintegration,

N = `(3.7×10^10)/0.6931` × 1600 × 365 × 24 × 60 × 60

= `1.8669/0.6931` × 10²¹

= 2.6936 × 10²¹

According to Avagadro’s principle,

6.023 × 10²³ atoms = 226 gm of radium

`"dN"/"dt"`= λN

N = `"dN"/"dt" 1/lambda`

`"dN"/"dt"` = 1 Curie

= 3.7 × 10¹⁰ disintegration per second

λ = 0.6931/("T"_1/2) per year

λ = `0.6931/(1600 xx 365 xx 24 xx 60 xx 60)` Per second × 2.6936 × 10²¹ atoms

`= 226/(6.023 xx 10^23) xx 2.6936 xx 10^21`

= 101.0715 × 10^-2 g

= 1.0107 gram

The activity of 1 curie of 1 gram of radium is approximately 1 gram.