A source contains two species of phosphorous nuclei, \\ce{_15^32P}\ (T1/2 = 14.3 d) and \\ce{_15^33P}\ (T1/2 = 25.3 d). - Physics

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Numerical

A source contains two species of phosphorous nuclei, \[\ce{_15^32P}\] (T1/2 = 14.3 d) and \[\ce{_15^33P}\] (T1/2 = 25.3 d). At time t = 0, 90% of the decays are from \[\ce{_15^32P}\]. How much time has to elapse for only 15% of the decays to be from \[\ce{_15^32P}\]?

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Solution

Data: \[\ce{_15^32P}\] : T1/2 = 14.3 d

∴ `lambda_1 = 0.693/(14.3 "d") = 0.04846  "d"^-1`

\[\ce{_15^32P}\] : T1/2 = 25.3 d

∴ `lambda_2 = 0.693/(25.3 "d") = 0.02739  "d"^-1`

At time t = 0, `("N"_"O1" lambda_1)/("N"_"O2"lambda_2) = (90%)/(10%) = 9`  ...(1)  and

at time t, `("N"_"O1" lambda_1"e"^(-lambda_1"t"))/("N"_"O2" lambda_2"e"^(-lambda_2"t")) = (15%)/(85%) = 3/17`  ...(2)

Dividing Eq. (1) by Eq. (2), we get,

`("N"_"O1" lambda_1)/("N"_"O2"lambda_2) * ("N"_"O1" lambda_1"e"^(-lambda_1"t"))/("N"_"O2" lambda_2"e"^(-lambda_2"t")) = 9/(3//17) = 153/3`

∴ `"e"^((lambda_1 - lambda_2)"t") = 153/3`

∴ `(lambda_1 - lambda_2)"t" = 2.303 log_10(153/3) = 2.303(log_10 153 - log_10 3)`

∴ (0.04846 - 0.02739) t = 2.303 (2.1847 - 0.4771)

∴ t = `((2.303)(1.7076))/0.02107` = 186.6 days

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Chapter 15: Structure of Atoms and Nuclei - Exercises [Page 343]

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Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 21 | Page 343

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