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Question
The disintegration rate of a radio-active sample is 1010 per hour at 20 hours from the start. It reduces to 5 × 109 per hour after 30 hours. Calculate the decay constant.
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Solution 1
Given:
A(t1) = 1010 per hour, where t1 = 20 h,
A(t2) = 5 × 109 per hour, where t2 = 30 h,
To find: λ = ?
Formula:
A(t) = `A_0e^{-lambdat}`
∴ `(A(t_1))/(A(t_2)) = (A_0e^{-lambdat_1})/(A_0e^{-lambdat_2}) = (e^{-lambdat_1})/(e^{-lambdat_2})`
∴ `10^10/(5 xx 10^-9) = (e^{-lambda20})/(e^{-lambda30}) = e^{lambda(30 - 20)} = e^{10lambda}`
∴ 2 = `e^{10lambda}`
∴ 2 = `e^{10lambda}`
∴ log2 = 10λloge = 10λ ..............(∵ loge = 1)
∴ 0.693 = 10λ
∴ λ = 0.0693
Solution 2
`N = N_0e^-(lambdat)`
`N_2/N_1 = e^(-lambda(t_2-t_1)`
Given:
N1 = 1.0 × 1010 per hour at t1 = 20 hours
N2 = 5.0 × 109 per hour at t2 = 30 hours
`N_2/N_1 = e^(-lambda(t_2-t_1)) =>(5xx10^9)/(1xx10^10) = e^(-lambda(10)) => 1/2 = e^(-10lambda)`
Take natural log
`(1/2) = -10lambda => -ln 2 = -10lambda => lambda = ln2/10 = 0.693/10`
= 0.0693 per hour
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