Chapter 32, Problem 6TP

To determine

(a)

To explain:

The equation of helium fusion reaction

Answer to Problem 6TP

The fusion reaction of helium is given by the following equation −

  $\left(\text{H}\right)\text{+}\left(\text{H}\right)\to \left(\text{H}\right)+\left(\text{n}\right)\text{+Energy}$

Explanation of Solution

Introduction:

The formation of one or more different atomic nuclei and the subatomic particles by the combination of two or more atomic nuclei under the chemical reaction which is known as the nuclear fusion.

The following are used to achieve the fusion −

1. Thermonuclear fusion
2. Inertial confinement fusion
3. Beam-beam fusion (Beam -target fusion)
4. Inertial electrostatic confinement
5. Muon − catalyzed fusion

There are two types of fusion reaction −

1. First reaction in which the preservation of the neutrons and protons number is available

  $\text{D+T}\to \text{He+n}$

Where,

Before the reaction − two protons and three neutrons present at left side, same present at right side also

2. In second type of reaction − conversion between the protons and neutrons
The initiation of the star burning available and formation of deuterium by the fusion of two hydrogen nuclei

  $\text{H+H}\to {\text{D+v+β}}^{\text{+}}$

Where,

Positron − ${\text{β}}^{\text{+}}$

Neutrinos - $\text{v}$

Hence, the fusion reaction for helium production is given by −

  $\left(\text{H}\right)\text{+}\left(\text{H}\right)\to \left(\text{H}\right)+\left(\text{n}\right)\text{+Energy}$

Conclusion:

In nuclear reaction, the energy is released when the mass of resultant particles is lesser than the initial reactant's mass.

To determine

(b)

To calculate:

The difference between the mass of reactants and mass of products is $\text{ 0}\text{.00304u}$

Answer to Problem 6TP

Explanation of Solution

Given:

  $\begin{array}{l}{\text{m}}_{\text{deuterium}}\text{=2}\text{.014102u}\\ {\text{m}}_{\text{H}\text{e}}\text{=3}\text{.01603u}\\ {\text{m}}_{\text{n}}\text{=1}\text{.008665u}\end{array}$

Formula used:

1. The mass of reactants is given by −

  ${\text{m}}_{\text{reactants}}{\text{= m}}_{\text{deuterium}}{\text{+m}}_{\text{deuterium}}$

Where,

The mass of deuterium − ${\text{m}}_{\text{deuterium}}$

2. The total mass of products is given by −

  ${\text{m}}_{\text{products}}{\text{= m}}_{\text{H}\text{e}}{\text{+m}}_{{}_{\text{n}}}$

Where,

The mass of the helium atom − ${\text{m}}_{\text{H}\text{e}}$

The mass of neutron − ${\text{m}}_{{}_{\text{n}}}$

3. The difference between the mass of reactant and products is given by −

  ${\text{Δm = m}}_{\text{reactants}}{\text{- m}}_{\text{products}}$

Calculation:

The mass of parent nuclei is given by −

  ${\text{m}}_{\text{reactants}}{\text{= m}}_{\text{deuterium}}{\text{+m}}_{\text{deuterium}}$

On putting the given values

  ${\text{m}}_{\text{reactants}}\text{= 2}\text{.014102+2}\text{.014102}$

  ${\text{m}}_{\text{reactants}}\text{= 4}\text{.028204u}$ ................... $\left(1\right)$

Now we can calculate the total mass of the products as −

  ${\text{m}}_{\text{products}}{\text{= m}}_{\text{H}\text{e}}{\text{+m}}_{{}_{\text{n}}}$

On putting the values from the given data

  ${\text{m}}_{\text{products}}\text{=3}\text{.01603+1}\text{.008665}$

  ${\text{m}}_{\text{products}}\text{=4}\text{.0247u}$ ................... $\left(2\right)$

Hence, by using the calculated mass of reactant and product, we are able to calculate difference between their masses −

  ${\text{Δm = m}}_{\text{reactants}}{\text{- m}}_{\text{products}}$

Put the calculated mass values from eq. $\left(1\right)$ and $\left(2\right)$ -

  $\begin{array}{l}{\text{Δm = m}}_{\text{reactants}}{\text{- m}}_{\text{products}}\\ \text{Δm = 4}\text{.028204u - 4}\text{.0247u}\\ \text{Δm = 0}\text{.00304u}\end{array}$

Conclusion:

From the above calculations,

The difference between the mass of reactants and mass of products is calculated as $\text{ 0}\text{.00304u}$

To determine

(c)

To calculate:

The energy produced from fusion reaction

Answer to Problem 6TP

The energy produced in fusion reaction is $\text{2}\text{.838MeV}$

Explanation of Solution

Given:

Speed of air − ${\text{c = 3×10}}^{\text{8}}\text{m/s}$

Mass difference between reactant and product - $\text{Δm = 0}\text{.00304u}$

Formula Used:

The energy released from fusion reaction is given by −

  ${\text{E = Δmc}}^2$

Where,

Speed of air − $\text{c }$

Mass difference between reactant and product - $\text{Δm }$

Calculation:

We know that −

  $\text{1u =1}{\text{.66×10}}^{\text{-27}}\text{kg}$

The mass difference is converted to kg −

  $\begin{array}{l}\text{Δm = 0}\text{.00304u}\\ \text{Δm =}\left(\text{0}\text{.00304×1}{\text{.66×10}}^{\text{-27}}\right)\text{kg}\\ \text{Δm = 5}{\text{.046×10}}^{\text{-27}}\text{kg}\end{array}$

Now we can calculate the produced energy by the fusion reaction −

  $\begin{array}{l}{\text{E = Δmc}}^{\text{2}}\\ \text{E =}\left(\text{5}{\text{.046×10}}^{\text{-27}}\text{kg}\right){\left({\text{3×10}}^{\text{8}}\text{m/s}\right)}^{\text{2}}\\ \text{E = 4}{\text{.541×10}}^{\text{-13}}\text{J}\end{array}$

Conclusion:

Hence, the energy produced from the fusion reaction process is calculated as $\text{4}{\text{.541×10}}^{\text{-13}}\text{J}$