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The half-life of 35S is 87.8 d. What percentage of 35S sample remains after 180 d?

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Question

The half-life of 35S is 87.8 d. What percentage of 35S sample remains after 180 d?

Numerical
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Solution

Given: t1/2 = 87.8 d, N0 = 100, t = 180 d

To find: % of 35S that remains after 180 days

Formulae:

  1. `lambda = 0.693/("t"_(1//2))`
  2. `lambda = 2.303/"t" log_10 ("N"_0/"N")`

Calculation:

  1. `lambda = 0.693/("t"_(1//2)) = 0.693/(87.8  "d") = 7.893 xx 10^-3 "d"^-1`
  2. Now, `lambda = 2.303/"t" log_10 ("N"_0/"N")` 

`log_10 ("N"_0/"N") = (lambda"t")/2.303`

`= (7.893 xx 10^-3 xx 180)/2.303`

= 0.617

Taking antilog on both sides we get,

`"N"_0/"N" = 4.140`

N = `100/4.140`

= 24.155 ≈ 24.2%

Percentage of 35S that remains after 180 d is 24.2%.

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Chapter 13: Nuclear Chemistry and Radioactivity - Exercises [Page 203]

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Balbharati Chemistry [English] Standard 11 Maharashtra State Board
Chapter 13 Nuclear Chemistry and Radioactivity
Exercises | Q 4. (B) | Page 203

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