A radioactive sample has activity of 10,000 disintegrations per second (dps) after 20 hours. After next 10 hours its activity reduces to 5,000 dps. Find out its half life and initial activity.
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Hey mate,
● Answer-
λ = 1.92×10^-5
t = 10 hrs
● Explaination-
# Given-
A1 = 10000 dps
A2 = 5000 dps
t1 = 20 hr = 72000 s
t2 = 10 hr = 36000 s
# Solution-
For a radioactive nucleus-
A1 = A e^(-λt1)
10000 = A e^(-72000λ)
A2 = A e^(-λt2)
5000 = A e^(-36000λ)
Dividing A1/A2,
10000 / 5000 = [Ae^(-72000λ)] / [Ae^(-36000λ)]
2 = e^(-72000λ) / e^(-36000λ)
2 = e^(-36000λ)
Taking log on both sides-
ln(2) = -36000λ ln(e)
λ = ln(2)/36000
λ = 1.92×10^-5
Half life is given by-
t = ln(2)/λ
t = 10 hrs
Initial activity is 1.92×10^-5 and half life of the radioactive nucleus is 10 hrs.
Hope this helps...
● Answer-
λ = 1.92×10^-5
t = 10 hrs
● Explaination-
# Given-
A1 = 10000 dps
A2 = 5000 dps
t1 = 20 hr = 72000 s
t2 = 10 hr = 36000 s
# Solution-
For a radioactive nucleus-
A1 = A e^(-λt1)
10000 = A e^(-72000λ)
A2 = A e^(-λt2)
5000 = A e^(-36000λ)
Dividing A1/A2,
10000 / 5000 = [Ae^(-72000λ)] / [Ae^(-36000λ)]
2 = e^(-72000λ) / e^(-36000λ)
2 = e^(-36000λ)
Taking log on both sides-
ln(2) = -36000λ ln(e)
λ = ln(2)/36000
λ = 1.92×10^-5
Half life is given by-
t = ln(2)/λ
t = 10 hrs
Initial activity is 1.92×10^-5 and half life of the radioactive nucleus is 10 hrs.
Hope this helps...
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