Chapter 21, Problem 21.70QP

(a)

Interpretation Introduction

Interpretation: For the given cases number of moles amount of heat released, energy released for each cases for both decay should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,$\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

Answer to Problem 21.70QP

The energy released in the above two decays is $\text{5}{\text{.59×10}}^{\text{-15 }}\text{J and 2}{\text{.84×10}}^{\text{-13}}\text{J}\text{.}$

The total amount of energy released is:

$\left(\text{5}{\text{.59×10}}^{\text{-15 }}\text{J}\right)\text{+2}{\text{.84×10}}^{\text{-13}}\text{J=2}{\text{.90×10}}^{\text{-13}}\text{J}$

Explanation of Solution

In case of ${\text{Sr}}^{\text{90}}$ decay, the mass defect is

$\text{Δm=}\left({\text{massY}}^{\text{90}}{\text{+mass e}}^{\text{-}}\right){\text{-massSr}}^{\text{90}}$

$\text{=}\left[\left(\text{89}\text{.907152amu+5}{\text{.4857×10}}^{\text{-4}}\text{amu}\right)\text{-89}\text{.907738amu}\right]\text{=-3}{\text{.7430×10}}^{\text{-5}}\text{amu}$

$\text{=}\left(\text{-}\right)\text{3}{\text{.7430× 10}}^{\text{-5}}\text{ amu)×}\frac{\text{1g}}{\text{6}{\text{.022×10}}^{\text{23}}\text{amu}}\text{ =6}{\text{.216 10}}^{\text{-29}}\text{ g =-6}{\text{.216×10}}^{\text{-32}}\text{kg}$

The energy change is given by :

$\text{ΔE}\left(\text{Δm}\right){\text{c}}^{\text{2}}$

$\left(\text{-6}{\text{.216×10}}^{\text{-32}}\text{kg}\right){\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}$

$\begin{array}{l}\text{=5}{\text{.59×10}}^{\text{-15}}{\text{kgm}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{=5}{\text{.59×10}}^{\text{-15}}\text{J}\end{array}$

Similarly for ${\text{Y}}^{\text{90}}$ decay, we have

$\text{Δm=}\left({\text{massZr}}^{\text{90}}{\text{+masse}}^{\text{-}}\right){\text{-massY}}^{\text{90}}$

$\text{[(89}\text{.904703 amu + 5}{\text{.4857 ×10}}^{\text{-4}}\text{amu-89}\text{.907152amu]=-1}{\text{.9004×10}}^{\text{-3}}\text{amu}$

$\left(\text{-1}\text{.9004 }\times {\text{10}}^{-3}\text{ amu}\right)\times \frac{\text{1 g}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ amu}}=-\text{3}\text{.156}\times {\text{ 10}}^{-27}\text{g}$

=$\text{- 3}\text{.156}\times {\text{ 10}}^{-30}\text{kg}$

And the energy change is given by:

$\text{ΔE=}$$\left(\text{-3}{\text{.156×10}}^{\text{-30}}\text{kg}\right){\text{(3}{\text{.00×10}}^{\text{8}}\text{m/s)}}^{\text{2}}\text{=-2}{\text{.84×10}}^{\text{-13}}\text{J}$

The energy released in the above two decays is $\text{5}{\text{.59×10}}^{\text{-15 }}\text{J and 2}{\text{.84×10}}^{\text{-13}}\text{J}\text{.}$

The total amount of energy released is:

$\left(\text{5}{\text{.59×10}}^{\text{-15 }}\text{J}\right)\text{+2}{\text{.84×10}}^{\text{-13}}\text{J=2}{\text{.90×10}}^{\text{-13}}\text{J}$

(b)

Interpretation Introduction

Interpretation: For the given cases number of moles amount of heat released, energy released for each cases for both decay should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,$\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

To calculate: The number of moles ${}_{}^{\text{90}}\text{Sr}$ decaying in a year.

Answer to Problem 21.70QP

The number of moles of nuclei which decay in a year is :

$\text{(1- 0}\text{.9756) mol = 0}\text{.0244 mol}$

This is a reasonable number since it takes 28.1 years for 0.5 moles of ${}_{}^{\text{90}}\text{Sr}$ to decay

Explanation of Solution

This calculation requires that we know the rate constant for the decay.

From half-life, we can calculate k.

$\text{k = }\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}\text{=}\frac{\text{0}\text{.693}}{\text{28}\text{.1yr}}\text{=0}{\text{.0247 yr}}^{\text{-1}}$

To calculate the number of moles ${}_{}^{\text{90}}\text{Sr}$ decaying in a year, we apply the following equation:

$\text{ln }\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=-kt}$

$\text{ln}\frac{\text{x}}{\text{1}}\text{=-}\left(\text{0}{\text{.0247yr}}^{\text{-1}}\right)\left(\text{1yr}\right)$

Where x is the number of moles nuclei left over .solving, we obtained the value of x

$\text{x = 0}{\text{.9756 mol}}^{\text{ 90}}\text{Sr}$

Thus the number of moles of nuclei which decay in a year is :

$\text{(1- 0}\text{.9756) mol = 0}\text{.0244 mol}$

This is a reasonable number since it takes 28.1 years for 0.5 moles of ${}_{}^{\text{90}}\text{Sr}$ to decay

(c)

Interpretation Introduction

Interpretation: For the given cases number of moles amount of heat released, energy released for each cases for both decay should be determined.

Concept Introduction:

• Energy required to break the nucleus into its corresponding proton and neutron is called nuclear binding energy
• This quantity represents the conversion of mass to energy occurs during an exothermic reaction.
• Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,$\Delta E=\left(\Delta m\right)c^2$

  Where, $\left(\Delta m\right)$ is called mass defect.

• The difference between mass of an atom and the sum of the masses of its proton, electron, and neutron is called Mass defect

To calculate: The heat released from 1 mole of ${}_{}^{\text{90}}\text{Sr}$ waste in a year.

Answer to Problem 21.70QP

The heat released from 1 mole of ${}_{}^{\text{90}}\text{Sr}$ waste in a year is given by:

$\text{heat released =}\left(\text{1}{\text{.47× 10}}^{\text{22}}\text{nuclei}\right)\text{×}\frac{\text{2}{\text{.90×10}}^{\text{-13}}\text{J}}{\text{1nucleus}}\text{=4}{\text{.26×10}}_{}^{\text{9}}\text{J=4}{\text{.26×10}}_{}^{\text{9}}\text{kJ}$

Explanation of Solution

The half –life of ${\text{Y}}^{\text{90}}$ is much shorter than that of ${}_{}^{\text{90}}\text{Sr}$.

We can assume that ${\text{Y}}^{\text{90}}$ formed from ${}_{}^{\text{90}}\text{Sr}$ and will be converted to ${\text{Z}}^{\text{90}}$.

The energy changes in (a) refer to decay of individual nuclei.

In 0.0244 moles, the number of nuclei that have decay is:

$\text{0}\text{.0244 mol×}\frac{\text{6}{\text{.022 ×10 }}^{\text{23}}\text{nuclei}}{\text{1mol}}\text{=1}{\text{.47×10}}^{\text{22}}\text{nuclei}$

There are two decay processes occurring, so we need to add the energy released for each process calculated in part (a).

So the heat released from 1 mole of ${}_{}^{\text{90}}\text{Sr}$ waste in a year is given by:

$\text{heat released =}\left(\text{1}{\text{.47× 10}}^{\text{22}}\text{nuclei}\right)\text{×}\frac{\text{2}{\text{.90×10}}^{\text{-13}}\text{J}}{\text{1nucleus}}\text{=4}{\text{.26×10}}_{}^{\text{9}}\text{J=4}{\text{.26×10}}_{}^{\text{9}}\text{kJ}$