Chapter 27, Problem 58QAP

To determine

(a)

The energy released in fusion reaction, $H+H\to He+n$

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  $\begin{array}{l}2+3=4+x\\ x=1\end{array}$

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  $H+H\to He+n$

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $H=2.01355321270u$

Atomic mass of $H=3.01604928199u$

Atomic mass of $He=4.002602u$

  $H+H\to He+n$

Mass defect

  $\begin{array}{l}\Delta m=\left(2.01355321270u+3.01604928199u\right)-\left(4.002602u+1.008664917u\right)\\ =0.018335577u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.018335577u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =17.07948MeV\end{array}$

Conclusion:

The energy released in fusion reaction is $17.07948MeV$.

To determine

(b)

The energy released in fusion reaction, $H+He\to Be+n$

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  $\begin{array}{l}4+4=7+x\\ x=1\end{array}$

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  $H+He\to Be+n$

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $Be=7.0169u$

Atomic mass of $H=4.03188u$

Atomic mass of $He=4.002602u$

  $H+He\to Be+n$

Mass defect

  $\begin{array}{l}\Delta m=\left(4.03188u+4.002602u\right)-\left(7.0169u+1.008664917u\right)\\ =0.008917083u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.008917083u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =8.306209MeV\end{array}$

Conclusion:

The energy released in fusion reaction is $8.306209MeV$.

To determine

(c)

The energy released in fusion reaction, $H+H\to He+n$

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  $\begin{array}{l}2+2=3+x\\ x=1\end{array}$

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  $H+H\to He+n$

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $H=2.01355321270u$

Atomic mass of $He=3.0160293u$

  $H+H\to He+n$

Mass defect

  $\begin{array}{l}\Delta m=\left(2.01355321270u+2.01355321270u\right)-\left(3.0160293u+1.008664917u\right)\\ =0.0024122084u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.0024122084u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =2.246958MeV\end{array}$

Conclusion:

The energy released in fusion reaction is $2.246958MeV$.

To determine

(d)

The energy released in fusion reaction, $H+H\to \gamma +X$

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  $\begin{array}{l}2+1=0+x\\ x=3\end{array}$

And

  $\begin{array}{l}1+1=0+y\\ x=2\end{array}$

So, the atomic number is 2.

Hence, the complete reaction would be

  $H+H\to \gamma +He$

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $H=2.01355321270u$

Atomic mass of $He=3.0160293u$

  $H+H\to \gamma +He$

Mass defect

  $\begin{array}{l}\Delta m=\left(2.01355321270u+1.007825u\right)-\left(3.0160293u\right)\\ =0.0053489127u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.0053489127u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =\text{4}\text{.9825}\text{MeV}\end{array}$

Conclusion:

The energy released in fusion reaction is $\text{4}\text{.9825}\text{MeV}$

To determine

(e)

The energy released in fusion reaction, $H+H\to H+n$

Explanation of Solution

To calculate the no of neutrons, release from the above equation. The mass should be equal to the left side and right side of the equation for balancing the equation. So, equate the left side and right side of the masses. We get,

  $\begin{array}{l}2+2=3+x\\ x=1\end{array}$

So, 1 neutron will be released from the above equation

Hence, the complete reaction would be

  $H+H\to H+n$

Mass of a proton $M_p=1.672621777\times {10}^{-27}kg=1.007276467u$

Mass of Hydrogen atom $M(H)=1.007825u$

Mass of a neutron $M_n=1.674927351\times {10}^{-27}kg=1.008664917u$

Atomic mass of $H=2.01355321270u$

Atomic mass of $He=3.0160293u$

  $H+H\to H+n$

Mass defect

  $\begin{array}{l}\Delta m=\left(2.01355321270u+2.01355321270u\right)-\left(3.0160293u+1.008664917u\right)\\ =0.0024122084u\end{array}$

Converting this into energy by below equation

  $E=m{c}^2$

  $\begin{array}{l}E=0.0024122084u\times 931.494061\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$u$}\right.\\ =2.246958MeV\end{array}$

Conclusion:

The energy released in fusion reaction is $2.246958MeV$