Chapter 30, Problem 87A

To determine

To calculate:

Complete fusion reaction for deuterium and energy released during the reaction.

Answer to Problem 87A

Complete fusion reaction for deuterium is $p+n\to H$ .

The energy released during the reaction is $-2.001958\times {10}^{17}\text{MeV}$

Explanation of Solution

Given:

  $\begin{array}{l}{m}_{atom}\left(H\right)=2.014102\text{u}\\ m\left(H\right)\text{=1}\text{.007825}\text{u}\\ m_n\text{=1}\text{.008665}\text{u}\end{array}$

Formula Used:

Number of neutrons is given as,

  $A-Z$

Where,

  $A:$ Atomic mass number

  $Z:$ Mass number Mass defect is given by formula,

  $\Delta m=\left[Zm\left(H\right)+\left(A-Z\right)m_n\right]-m_{atom}$

Where,

  $\Delta m:$ Mass defect

  $m\left(H\right):$ Mass of hydrogen

  $m_n:$ Mass of neutron

  $m_{atom}:$ Mass of overall atom

  $A:$ Atomic mass number

  $Z:$ Atomic number

Energy is given by formula,

  $E=\Delta m\times c^2$

Where,

  $\Delta m:$ Mass defect

  $c:$ Speed of light

Calculation:

To obtain the number of protons,

As the number of protons in the nucleus of the atom is equal to the atomic number $\text{Z}$ ,

Number of protons $\text{=1}$

Number of neutrons is given by subtracting atomic number form the mass number,

  $\text{A-Z}$

Substitute the values in above formula,

  $\begin{array}{l}\text{=2-1}\\ \text{=1}\end{array}$

Number of neutrons $\text{=1}$

To obtain energy, firstly there is need to find mass defect.

Mass effect can be calculated by formula,

Substituting the values in above formula,

  $\begin{array}{l}\Delta m=\left[\left(1\times 1.007825u\right)+\left(1\times 1.008665u\right)\right]-2.014102\text{u}\\ =2.01649\text{u}-2.014102\text{u}\\ \text{=-0}\text{.002388}\text{u}\end{array}$

Energy released during the reaction can be calculated by formula,

  $E=\Delta m\times c^2$

Substituting the values in above formula,

  $\begin{array}{l}E=\left(\text{-0}\text{.002388}\text{u}\right)\times {\left(3\times {10}^8\right)}^2\\ =-2.1492\times {10}^{14}\text{u}\end{array}$

Converting the value of energy in $\text{MeV}$ ,

As, $\text{1u=931}\text{.49}\text{MeV}$ ,

  $\begin{array}{l}E=\left(-2.1492\times {10}^{14}\text{u}\right)\times \left(931.49\text{MeV}\right)\\ =-2.001958\times {10}^{17}\text{MeV}\end{array}$

Conclusion:

Hence, complete fusion reaction for deuterium is $p+n\to H$ and the energy released during the reaction is $-2.001958\times {10}^{17}\text{MeV}$ .