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?

Answer 1

a) average life ( τ) = = 2.04* 10¹⁷ s .  

b) half-life = 1.414*10¹⁷ s .

c) activity has decreased by the factor of 4.014 . i,e activity has become 0.249 of initial.

Explanation:

a) average life ( τ) = 1 / λ          where λ   is the decay constant

                         = 1/ 4.9 × 10−18 S−1.

                        = 0.204 * 10¹⁸ s

                        = 2.04* 10¹⁷ s .

b) half life t1/2 = 0.693 / λ

                     = 1.414*10¹⁷ s .  

c) A(t) = A(0) e⁻λt  

where A(t) is activity at time t   and  A(0) is activity at time 0

   A(t) /A(0) =  e⁻λt

                 = e⁻(4.9 × 10−18 S−1 * 9 × 10⁹ * 365*24*60*60 )

be careful to convert years into second while calculating

                = e⁻1.39

               = 1/4.014 or 0.249

Answer 2

Explanation:

It is given:

Constant of decay, λ =$\lambda=4.9 \times 10^{-18} \mathrm{s}^{-1}$

(a) Uranium has the average life, τ, which is shown as

$\tau=1 \lambda=14.9 \times 1018 \mathrm{s}$

$=24 \times 36 \text { years } \times 10164.9 \times 365=6.47 \times 105$years

(b) Uranium has the half-life, T12 shown as

$\mathrm{T} 12=0.693 \lambda=0.6934 .9 \times 10-18$

$=0.6934 .9 \times 1018 \mathrm{s}=0.1414 \times 1018 \mathrm{s}=4.5 \times 109 \mathrm{years}$

(c) Time is denoted as $t=9 \times 10^{9} \text { years }$

Sample’s s activity (A), at any time t, is shown as

$A=A O 2 t T 1 / 2$

where, A0 = sample’s activity at t = 0

Therefore, $A O A=29 \times 1094.5 \times 109=4$