Chapter 19, Problem 41QAP

Interpretation Introduction

Interpretation:

Interpret the most stable isotope and determine the hottest isotope. Interpret the amount of following isotope left after 24 hr.

Kr-73 with half life 27 s

Kr-74 with half life 11.5 min

Kr-76 with half life 14.8 h

Kr-81 with half life $\text{2}\text{.1 }\times {\text{ 10}}^{\text{5}}\text{yr}$.

Concept Introduction:

The time elapsed can be identified through the below formula:

$\left({\text{N /N}}_{\text{0}}\right)\text{ = }{\left(\text{1 /2}\right)}^{\text{t /t half}}$

N is number of mole left after time t

N₀ is initial number of mole

T_{h a l f} is half life.

Answer to Problem 41QAP

Half life of Kr-81 is greatest of all that is $\text{2}\text{.1 }\times {\text{ 10}}^{\text{5}}\text{yr}$ and thus is more stable while half life of Kr-73 is lowest of all that is 27 s and thus is less stable.

Explanation of Solution

Higher the half life, more the molecule take time to reduce to half of its concentration and is more stable that’s why it is taking more time to reduce.

Half life of Kr-81 is greatest of all that is $\text{2}\text{.1 }\times {\text{ 10}}^{\text{5}}\text{yr}$ and thus is more stable.

Hottest isotope have lower half life and react or disintegrate easily because of low stablity. Half life of Kr-73 is lowest of all that is 27 s and thus is less stable.

$\left({\text{N /N}}_{\text{0}}\right)\text{ = }{\left(\text{1 /2}\right)}^{\text{t /t half}}$

Calculate N for each isotope

N ₀ for each isotope = 125 micro gram

T half is given

T = 24 hr

For Kr-73

$\begin{array}{l}\begin{array}{l}{\text{T}}_{\text{half}}\text{= 27 s}\hfill \\ \text{t = 24 h = 24 }\times \text{ 3600 = 86400 sec}\text{.}\hfill \\ {\text{N}}_{\text{0}}\text{= 125 µg}\hfill \\ {\text{N /N}}_{\text{o}}\text{= }{\left(\text{ 1/2}\right)}^{\text{t / t half}}\hfill \\ \text{N /125 = }{\left(\text{1 /2}\right)}^{\text{86400 /27}}\hfill \\ \text{N = 125 }{\left(\text{1 /2}\right)}^{\text{3200}}\hfill \\ \hfill \end{array}\\ \text{N = 8}\text{.36 }\times {\text{ 10}}^{\text{-14}}\text{µg}\end{array}$

For Kr-74

$\begin{array}{l}\begin{array}{l}{\text{T}}_{\text{half}}\text{= 27 s}\hfill \\ \text{t = 11}\text{.5 min = 11}\text{.5 }\times \text{ 60 = 690 sec}\text{.}\hfill \\ {\text{N}}_{\text{0}}\text{= 125 µg}\hfill \\ {\text{N /N}}_{\text{o}}\text{= }{\left(\text{ 1/2}\right)}^{\text{t / t half}}\hfill \\ \text{N /125 = }{\left(\text{1 /2}\right)}^{\text{690 /27}}\hfill \\ \text{N = 125 }{\left(\text{1 /2}\right)}^{\text{25}\text{.5}}\hfill \end{array}\\ \text{N = 12}\text{.72 }\times {\text{ 10}}^{\text{-4}}\text{µg}\end{array}$

For Kr-76

$\begin{array}{l}\begin{array}{l}{\text{T}}_{\text{half}}\text{= 27 s}\hfill \\ \text{t = 14}\text{.8 h = 14}\text{.8 }\times \text{ 3600 = 53280 sec}\text{.}\hfill \\ {\text{N}}_{\text{0}}\text{= 125 µg}\hfill \\ {\text{N /N}}_{\text{o}}\text{= }{\left(\text{ 1/2}\right)}^{\text{t / t half}}\hfill \\ \text{N /125 = }{\left(\text{1 /2}\right)}^{\text{53280 /27}}\hfill \\ \text{N = 125 }{\left(\text{1 /2}\right)}^{\text{1973}\text{.8}}\hfill \end{array}\\ \text{N = 4}\text{.28 }\times {\text{ 10}}^{\text{-8}}\text{µg}\end{array}$

For Kr-81

T_{h a l f} = 27 s

$\text{t = 2}\text{.1 }\times {\text{ 10}}^{\text{5}}\text{yr = 6}\text{.615 }\times {\text{ 10}}^{\text{12}}\text{sec}\text{.}$

N₀ = 125 µg

${\text{N /N}}_{\text{o}}\text{= }{\left(\text{ 1/2}\right)}^{\text{t / t half}}$

$\text{N/125=(1/2)6}{\text{.615*10}}^{\text{1}}\text{2/27}$

$\text{N=125(1/2)0}{\text{.25*10}}^{\text{1}}\text{2}$

$\text{N = 24}{\text{.14 \&*\#x00A0;10}}^{\text{-46}}\text{µg}$.

Conclusion

Hence, Half life of Kr-81 is greatest of all that is $\text{2}\text{.1 }\times {\text{ 10}}^{\text{5}}\text{yr}$ and thus is more stable while half life of Kr-73 is lowest of all that is 27 s and thus is less stable.