Chapter 46, Problem 23SP

(a)

To determine

The energy released when one atom of uranium undergoes the fission reaction, $n+\text{U}\to \text{B}\text{a}+\text{N}\text{b}+5n+5e$.

Answer to Problem 23SP

Solution:

$182\text{ MeV}$

Explanation of Solution

Given data:

The atomic mass of neutron is $1.0087\text{ u}$.

The atomic mass of uranium is $\text{235}\text{.0439 u}$.

The atomic mass of barium is $\text{137}\text{.9050 u}$.

The atomic mass of niobium is $92.9060\text{ u}$.

The atomic mass of electron is $0.00055\text{ u}$.

Formula used:

The expression for the mass loss is written as,

$\Delta m=\text{Reactant mass}-\text{Product mass}$

The expression for the energy released is written as,

$\Delta E=\left(\Delta m\right)c^2$

Here, $c$ is the speed of light.

Explanation:

Write the given reaction in the question.

$n+\text{U}\to \text{B}\text{a}+\text{N}\text{b}+5n+5e$

Write the expression for the reactant mass.

$\text{Reactant mass}=m_n+m_{\text{U}}$

Here, $m_n$ is the atomic mass of neutron and $m_{\text{U}}$ is the atomic mass of uranium.

Substitute $1.0087\text{ u}$ for $m_n$ and $\text{235}\text{.0439 u}$ for $m_{\text{U}}$

$\begin{array}{c}\text{Reactant mass}=1.0087\text{ u}+\text{235}\text{.0439 u}\\ =236.0526\text{ u}\end{array}$

Write the expression for the product mass.

$\text{Product mass}=m_{\text{B}\text{a}}+m_{\text{N}\text{b}}+5\left(m_n\right)+5\left(m_e\right)$

Here, $m_{\text{B}\text{a}}$ is the atomic mass of barium, $m_{\text{N}\text{b}}$ is the atomic mass of niobium, and $m_e$ is the atomic mass of electron.

Substitute $\text{137}\text{.9050 u}$ for $m_{\text{B}\text{a}}$, $92.9060\text{ u}$ for $m_{\text{N}\text{b}}$, $1.0087\text{ u}$ for $m_n$, and $0.00055\text{ u}$ for $m_e$

$\begin{array}{c}\text{Product mass}=\text{137}\text{.9050 u}+92.9060\text{ u}+5\left(1.0087\text{ u}\right)+5\left(0.00055\text{ u}\right)\\ =235.85725\text{ u}\end{array}$

Recall the expression for the mass loss.

$\Delta m=\text{Reactant mass}-\text{Product mass}$

Substitute $236.0526\text{ u}$ for $\text{Reactant mass}$ and $235.85725\text{ u}$ for $\text{Product mass}$

$\begin{array}{c}\Delta m=236.0526\text{ u}-235.85725\text{ u}\\ =0.19535\text{ u}\end{array}$

Recall the expression for the energy released.

$\Delta E=\left(\Delta m\right)c^2$

Substitute $0.19535\text{ u}$ for $\Delta m$

$\begin{array}{c}\Delta E=\left(0.19535\text{ u}\right)c^2\left(\frac{931.494095\text{ MeV}}{1\text{ u}{c}^2}\right)\\ =182\text{ MeV}\end{array}$

Conclusion:

Hence, the energy released when one atom of uranium undergoes the fission reaction is $182\text{ MeV}$.

(b)

To determine

The energy released when $1\text{ kg}$ of uranium atom undergoes the fission reaction,

$n+\text{U}\to \text{B}\text{a}+\text{N}\text{b}+5n+5e$.

Answer to Problem 23SP

Solution:

$7.5\times {10}^{13}\text{ J}$

Explanation of Solution

Given data:

The atomic mass of neutron is $1.0087\text{ u}$.

The atomic mass of uranium is $\text{235}\text{.0439 u}$.

The atomic mass of barium is $\text{137}\text{.9050 u}$.

The atomic mass of niobium is $92.9060\text{ u}$.

The atomic mass of electron is $0.00055\text{ u}$.

Formula used:

Molar mass of $\text{U-235 }$ is $238\text{ g}$.

The number of atoms in $238\text{ g}$ of uranium is $6.02\times {10}^{23}$ atoms.

Explanation:

Calculate the number of atoms present in $1\text{ kg}$ of uranium.

$n_{\text{U}}=\left(\frac{\text{number of atoms present in }238\text{ g of U-235 }}{\text{molar mass of U-235}}\right)\times 1000$

Substitute $6.02\times {10}^{23}$ for $\text{number of atoms present in }238\text{ g of U-235 }$ and $238\text{ g}$ for $\text{mass of U-235}$

$\begin{array}{c}{n}_{\text{U}}=\left(\frac{6.02\times {10}^{23}}{238\text{ g}}\right)\times 1000\\ =0.0253\times {10}^{26}\text{ per kg}\end{array}$

Thus, there are $0.0256\times {10}^{26}$ number of atoms present in $1\text{ kg}$ of uranium.

Write the expression for the energy released in $1\text{ kg of }\text{U}$.

$E_{\text{U}}=n_{\text{U}}{\left(\Delta E\right)}_{\text{ per atom}}$

Substitute $182\text{ MeV}$ for ${\left(\Delta E\right)}_{\text{ per atom}}$ and $0.0256\times {10}^{26}\text{ atoms}$ for $n_{\text{U}}$

$\begin{array}{c}{E}_{\text{U}}=\left(0.0256\times {10}^{26}\right)\left(182\text{ MeV}\right)\\ =\left(4.6592\times {10}^{26}\text{ MeV}\right)\left(\frac{{10}^6\text{ eV}}{1\text{ MeV}}\right)\\ =\left(4.6592\times {10}^{32}\text{ eV}\right)\left(\frac{1.6\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)\\ =7.5\times {10}^{13}\text{ J}\end{array}$

Conclusion:

Hence, the energy released when $1\text{ kg}$ of uranium atom undergoes the fission reaction is $7.5\times {10}^{13}\text{ J}$.