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