Chapter 19, Problem 3SP

(a)

To determine

The mass difference $\Delta m$ between the reactants and the products for the reaction ${}_{\text{1}}{\text{H}}^{\text{2}}\text{+}{}_{\text{1}}{\text{H}}^{\text{2}}\to {}_{\text{2}}{\text{He}}^{\text{3}}\text{+}{}_{\text{0}}{\text{n}}^{\text{1}}$.

Answer to Problem 3SP

The mass difference $\Delta m$ between the reactants and the products for the reaction ${}_{\text{1}}{\text{H}}^{\text{2}}\text{+}{}_{\text{1}}{\text{H}}^{\text{2}}\to {}_{\text{2}}{\text{He}}^{\text{3}}\text{+}{}_{\text{0}}{\text{n}}^{\text{1}}$ for the reaction is $3.51\times {10}^{-3}\text{u}$

Explanation of Solution

Given info: The mass of ${}_{\text{1}}{\text{H}}^{\text{2}}$ is $2.014102\text{u}$ , the mass of ${\text{He}}^3$ is $3.016029\text{u}$ and the mass of neutron is $1.008665$.

Write the expression for the mass difference for the reaction.

$\Delta m=2\left(m_{\text{H}}\right)-\left(m_{\text{He}}+m_{\text{n}}\right)$

Here,

$\Delta m$ is the mass difference

$m_{\text{He}}$ is the mass of the ${}_2{\text{He}}^3$

$m_{\text{H}}$ is the mass of the ${}_{\text{1}}{\text{H}}^{\text{2}}$

$m_{\text{n}}$ is the mass of the neutron

Substitute $3.016029\text{u}$ for $m_{\text{He}}$,$2.014102\text{u}$ for $m_{\text{H}}$ and $1.008665$ for $m_{\text{n}}$ in the above equation to get $\Delta m$.

$\begin{array}{c}\Delta m=2\left(2.014102\text{u}\right)-\left(3.016029\text{u}+1.008665\right)\\ =3.51\times {10}^{-3}\text{u}\end{array}$

Therefore, the mass difference between the reactants and products for the reaction is $3.51\times {10}^{-3}\text{u}$

Conclusion:

Thus, the mass difference $\Delta m$ between the reactants and the products for the reaction ${}_{\text{1}}{\text{H}}^{\text{2}}\text{+}{}_{\text{1}}{\text{H}}^{\text{2}}\to {}_{\text{2}}{\text{He}}^{\text{3}}\text{+}{}_{\text{0}}{\text{n}}^{\text{1}}$ for the reaction is $3.51\times {10}^{-3}\text{u}$

(b)

To determine

Energy unit corresponds to the mass difference.

Answer to Problem 3SP

The mass difference is equal to $5.25\times {10}^{-13}\text{J}$.

Explanation of Solution

Given info: The unified mass is equal to $1.661\times {10}^{-27}\text{kg}$ and velocity of light is light is equal to $3\times {10}^{-8}\text{m}/\text{s}$.

Write the mass-equivalence relation for the mass $\Delta m$.

$E=\Delta m{c}^2$

Here,

$E$ is the energy equivalent to the mass difference between the reactants and products

$\Delta m$ is the mass difference

$c$ is the velocity of light in vacuum

Convert $3.51\times {10}^{-3}\text{u}$ into kilogram.

$\begin{array}{c}3.51\times {10}^{-3}\text{u}\text{=}3.51\times {10}^{-3}\text{u}\times \frac{1.661\times {10}^{-27}\text{kg}}{1\text{u}}\\ =5.83\times {10}^{-30}\text{kg}\end{array}$

Substitute $5.83\times {10}^{-30}\text{kg}$ for $\Delta m$ , $3\times {10}^{-8}\text{m}/\text{s}$ for $c$ in the above equation to get $E$.

$\begin{array}{c}E=\left(5.83\times {10}^{-30}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ =5.25\times {10}^{-13}\text{J}\end{array}$

Conclusion:

Thus, the energy equivalent to that of mass difference between the reactants and products in the reaction ${}_{\text{1}}{\text{H}}^{\text{2}}\text{+}{}_{\text{1}}{\text{H}}^{\text{2}}\to {}_{\text{2}}{\text{He}}^{\text{3}}\text{+}{}_{\text{0}}{\text{n}}^{\text{1}}$ is $5.25\times {10}^{-13}\text{J}$.

(c)

To determine

Whether energy calculated is released. Determine where it goes.

Answer to Problem 3SP

The energy is released in the reaction and the released energy transformed in the form of kinetic energy of the products.

Explanation of Solution

Fusion reaction is the reaction in which two hydrogen atoms combined together to form heavier nucleus.

Write the fusion reaction mentioned in the text book.

${}_{\text{1}}{\text{H}}^{\text{2}}\text{+}{}_{\text{1}}{\text{H}}^{\text{2}}\to {}_{\text{2}}{\text{He}}^{\text{3}}\text{+}{}_{\text{0}}{\text{n}}^{\text{1}}$

Here, the total mass of the reactants is less than the total mass of the products. This difference in the mass converts as energy and transform in the form of kinetic energy of the Helium and neutron.

Therefore according conservation of energy, total energy should be a constant and one form of energy can be converted into any other form of energy. In order to conserve the energy the neutron particle moves with more kinetic energy.

Conclusion:

Thus, energy is released in the reaction and it is in the form of kinetic energy of the products.