The isotope \\ce{^57Co}\ decays by electron capture to \\ce{^57Fe}\ with a half-life of 272 d. The \\ce{^57Fe}\ nucleus is produced in an - Physics

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Answer in Brief

The isotope \[\ce{^57Co}\] decays by electron capture to \[\ce{^57Fe}\] with a half-life of 272 d. The \[\ce{^57Fe}\] nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
(a) Find the mean lifetime and decay constant for 57Co.
(b) If the activity of a radiation source 57Co is 2.0 µCi now, how many 57Co nuclei does the source contain?

c) What will be the activity after one year?

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Solution

Data: T1/2 = 272 d = 272 × 24 × 60 × 60s = 2.35 × 107 s, A0 = 2.0 µCi= 2.0 × 10-6 × 3.7 × 1010 = 7.4 × 104 dis/s

t = 1 year = 3.156 × 107 s

(a) `"T"_(1//2) = 0.693/lambda = 0.693 tau`

∴ The mean lifetime for 57Co =

`tau = "T"_(1//2)/0.693 = (2.35 xx 10^7)/0.693 = 3.391 xx 10^7`s

The decay constant for 57Co = `lambda = 1/tau`

`= 1/(3.391 xx 10^7"s")`

= 2.949 × 10-8 s-1

(b) `"A"_0 = "N"_0lambda`

∴ `"N"_0 = "A"_0/lambda = "A"_0tau`

`= (7.4 xx 10^4)(3.391 xx 10^7)`

= 2.509 × 1012 nuclei

(c) A(t) = `"A"_0"e"^(-lambda"t") = "2e"^(-(2.949 xx 10^-8)(3.156 xx 10^7))`

`= 2"e"^(-0.9307) = 2//"e"^0.9307`

Let x = `"e"^(0.9307)`

∴ logex = 0.9307

∴ 2.303 log10x = 0.9307

∴ `log_10"x" = 0.9307/2.303 = 0.4041`

∴ x = antilog 0.4041=2.536

∴ A(t) = `2/2.536 mu"Ci" = 0.7886 mu`Ci

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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 20 | Page 343

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