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प्रश्न
The decay constant of `""_80^197`Hg (electron capture to `""_79^197`Au) is 1.8 × 10−4 S−1. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?
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उत्तर
Given :-
Decay Constant of `""_80^197"Hg" , lambda = 1.8 xx 10^-4 "s"^-1`
(a)
Half-life, `T_"1/2" = 0.693/lambda`
`⇒ T_"1/2" = 0.693/(1.8 xx 10^-4)`
= 3850 s=64 minutes
(b)
Average life, `T_(av) = T_"1/2"/0.693`
`= 64/0.693`
= 92 minutes
(c)
Number of active nuclei of mercury at t = 0 = N0 = 100
Active nuclei of mercury left after conversion of 25% isotope of mercury into gold = N = 75
Now , `N/N_0 = e^(-lambda t)`
Here,
N = Number of inactive nuclei
`N_0` = Number of nuclei at t = 0
`lambda =` Disintegration constant
On substituting the values, we get
`75/100 = e^(-lambdat)`
`⇒ 0.75 = e^(-lambda x)`
`⇒ "In" 0.75 = - lambda t`
`⇒ t = ("In" 0.75)/-0.00018`
= 1600 s
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