1.

What weight of ""^(14)C (t_((1)/(2))=5760 years) will make one curie of it ? [one curie gives 3.7 xx 10^(10) disintegrations/second (dps)]

Answer»

Solution :Let the weight of `""^(14)C` be w gram Number of `""^(14)C` nuclei= mole `XX` Av. CONSTANT
`=(w)/(14) xx 6.022 xx 10^(23)`
and `t_((1)/(2))` in second `=5760 xx 365 xx 24 xx 60xx 60`
Now, we have, `-(d(N))/(dt)= lamda (N)= (0.6932)/(t_((1)/(2))) xx N`
`3.7 xx 10^(10)= (0.6932)/(5760 xx 365 xx 24 xx 60 xx 60)xx (w)/(14) xx 6.022 xx 10^(23)`
w= 0.225g


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