Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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The Decay Constant of 238u is 4.9 × 10−18 S−1. (A) What is the Average-life of 238u? (B) What is the Half-life of 238u? (C) by What Factor Does the Activity of a 238u Sample Decrease in 9 × 109 Years? - Physics

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Question

The decay constant of 238U is 4.9 × 10−18 S−1. (a) What is the average-life of 238U? (b) What is the half-life of 238U? (c) By what factor does the activity of a 238U sample decrease in 9 × 109 years?

Solution

Given:
Decay constant, `lambda = 4.9 xx 10^-18  "s"^-1`

(a) Average life of uranium (`tau`) is given by

`tau = 1/lambda`

= `1/(4.9 xx 10^-18)`

= `1/4.9 xx 10^18  "s"`

= `10^16/(4.9 xx 365 xx 24 xx 36)  "years"`

= `10^16/(4.9 xx 365 xx 24 xx 36)  "years"`

= `6.47 xx 10^-7 xx 10^16 "years"`

= `6.47 xx 10^9  "years"`

(b) Half-life of uranium (`T_"1/2"`) is given by 

`T_"1/2" = 0.693/lambda = 0.693/(4.9 xx 10^-18)`

= `0.693/4.9 xx 10^18  "s"`

= `0.1414 xx 10^18  "s"`

= `(0.1414 xx 10^18)/(365 xx 24 xx 3600)`

= `(1414 xx 10^12)/(365 xx 24 xx 36)`

= `4.48 xx 10^-3 xx 10^12`

= `4.5 xx 10^9` years

(c) Time, t = 9 × 109 years
 Activity (A) of the sample, at any time t, is given by

`A = A_0/2^(t/T_"1/2")`

Here ,  `A_0`  = Activity of the sample at t = 0

`therefore A_0/A = 2^((9 xx 10^9)/(4.5 xx 10^9)) = 2^2 = 4`

  Is there an error in this question or solution?

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Solution The Decay Constant of 238u is 4.9 × 10−18 S−1. (A) What is the Average-life of 238u? (B) What is the Half-life of 238u? (C) by What Factor Does the Activity of a 238u Sample Decrease in 9 × 109 Years? Concept: Radioactivity - Law of Radioactive Decay.
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