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प्रश्न
Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is :
\[\ce{^38Sulphur ->[half-life][= 2.48h] ^{38}Cl ->[half-life][= 0.62h] ^38Air (stable)}\]
Assume that we start with 1000 38S nuclei at time t = 0. The number of 38Cl is of count zero at t = 0 and will again be zero at t = ∞ . At what value of t, would the number of counts be a maximum?
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उत्तर
\[\ce{^38S ->[][2.48 h] ^38Cl ->[][0.62h] 38Ar}\]
At time t, Let 38S have N1(t ) active nuclei and 38Cl have N2(t) active nuclei.
`(dN_1)/(dt) = -λ_1N_1` = rate of formation of Cl38.
Also `(dN_2)/(dt) = - λ_1N_2 + λ_1N_1`
But `N_1 = N_0e^(-λ_1t)`
`(dN_2)/(dt) = - λ_1 N_0e^(-λ_1t) - λ_2N_2`
Multiplying by `e^(λ_2t) dt` and rearranging
`e^(λ_2t) dN_2 + λ_2N_2e^(λ_2t) dt = λ_1N_0e^((λ_2 - λ_1)t) dt`
Integrating both sides.
`N_2e^(λ_2t) = (N_0λ_1)/(λ_2 - λ_1) e^((λ_2 - λ_1)t) + C`
Since at t = 0, N2 = 0, C = `-(N_0λ_1)/(λ_2 - λ_1)`
∴ `N_2e^(λ_2t) = (N_0λ_1)/(λ_2 - λ_1) e^((λ_2 - λ_1)t) + C`
`N_2 = (N_0λ_1)/(λ_2 - λ_1) (e^(-λ.t) - e^(-λ_2t))`
For maximum count, `(dN_2)/(dt)` = 0
On solving, `t = (In λ_1/λ_2)/(λ_1 - λ_2)`
= In `(2.48/0.62)/(2.48 - 0.62)`
= `(In 4)/1.86`
= `(2.303 log 4)/1.86`
= 0.745 s.
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