# A sample of 32P initially shows activity of one Curie. After 303 days, the activity falls to 1.5 × 104 dps. What is the half-life of 32P? - Chemistry

Numerical

A sample of 32P initially shows activity of one Curie. After 303 days, the activity falls to 1.5 × 104 dps. What is the half-life of 32P?

#### Solution

Given: 1 Ci = 3.7 × 1010 dps,

("-dN"_0)/"dt" = 3.7 xx 10^10 "dps" and ("-dN"/"dt") = 1.5 xx 10^4 dps, t = 303 days

To find: t1/2

Formulae:

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

Calculation:

1. lambda = 2.303/"t" log_10 ("N"_0/"N")
lambda = 2.303/303 log_10 ((3.7 xx 10^10)/(1.5 xx 10^4))  ....[therefore (- "dN")/"dt" prop "N"]
= 0.04859 d-1
2. "t"_(1//2) = 0.693/lambda = 0.693/0.0459
= 14.27 d (by using log table)

Half-life of 32P is 14.27 days

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#### APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 13 Nuclear Chemistry and Radioactivity
Exercises | Q 4. (J) | Page 203