Question

The decay constant of Hg19780 (electron capture to Au19779) is 1.8 × 10−4 S−1. (a) What is the half-life? (b) What is the average-life? (c) How much time will it take to convert 25% of this isotope of mercury into gold?

Answer 1

Explanation:

It is given in the question:

$\mathrm{Hg} 80197$ has the decay constant, $\lambda=1.8 \times 10^{-4} \mathrm{s}-1$

(a) Half-life, T12 = 0.693λ

$\Rightarrow \mathrm{T} 1 / 2=0.6931 .8 \times 10-4$      

= 64 minutes

(b) Average life, Tav = T1/20.693                              

= 92 minutes

(c) Number of mercury’s active nuclei at t = 0

$=N_{0}$

= 100

After the conversion of 25% of mercury isotope into gold, the mercury’s active nuclei are N = 75

Now,

NN0 = e-λt

Here, N = Inactive nuclei’s number  

λ = Constant of disintegration constant

$N_{0}$ = at t = 0, number of nuclei  

When the values are substituted, we obtain

75100 = e-λt

⇒0.75 = e-λx

⇒t = In 0.75 - 0.00018        

=1600 s