Question

The ratio of the amount of energy released as a result of the fusion of $1\,kg$ hydrogen $\left( {{E_1}} \right)$ and fission of $1\,kg$ $_{92}{U^{235}}$ will be
A. $1.28$
B. $3.28$
C. $5.28$
D. $7.28$

Answer 1

Hint- The energy released when one uranium nucleus undergoes fission is
$E = 200\,MeV$
The number of atoms $n$ in $1\,kg$ uranium can be found out by finding the ratio of given mass to molecular mass and multiplying the ratio with Avogadro number.
$n = \dfrac{m}{M} \times {N_A}$
Where, Avogadro number, ${N_A} = 6.023 \times {10^{23}}$
The total energy released in fusion of $1\,kg$ uranium is the product of energy released in fission of one nucleus and the total number of nuclei in $1\,kg$.
${E_1} = n \times E$
$1\,MeV = 1.6 \times {10^{ - 13}}\,J$
Energy released during the fusion of $1\,kg$ hydrogen has a standard value. Let it be denoted as ${E_2}$.
${E_2} = 6.4 \times {10^{14}}\,J$

Step by step solution:
The energy released when one uranium nucleus undergoes fission is
$E = 200\,MeV$
We need to find the energy released when $1\,kg$ uranium undergoes fission. In order to find that we need to know the number of atoms in $1\,kg$ uranium.
The number of atoms $n$ in $1\,kg$ uranium can be found out by finding the ratio of given mass to molecular mass and multiplying the ratio with Avogadro number.
$n = \dfrac{m}{M} \times {N_A}$
Where, Avogadro number, ${N_A} = 6.023 \times {10^{23}}$
mass, $m = 1\,kg = 1000g$
Molecular mass of $_{92}{U^{235}}$, $M = 235$
Substituting these values in the equation, we get
Number of atoms,
$n = \dfrac{{1000}}{{235}} \times 6.023 \times {10^{23}} \\\ = 2.562 \times {10^{24}} \\\$
So, the total energy released in fusion of $1\,kg$ uranium is the product of energy released in fission of one nucleus and the total number of nuclei in $1\,kg$.
${E_1} = n \times E$
Therefore, we get the total energy as
${E_1} = 2.562 \times {10^{24}} \times 200\,MeV \\\ = 5.12 \times {10^{26}}\,MeV \\\$
We know that $1\,MeV = 1.6 \times {10^{ - 13}}\,J$
Therefore,
${E_1} = 5.12 \times {10^{26}} \times 1.6 \times {10^{ - 13}}\,J \\\ = 8.192 \times {10^{13}}\,J \\\$
Energy released during the fusion of $1\,kg$ hydrogen has a standard value. Let it be denoted as ${E_2}$.
${E_2} = 6.4 \times {10^{14}}\,J$
Now take the ratio of these two energies.
So, we get
$\dfrac{{{E_2}}}{{{E_1}}} = \dfrac{{6.4 \times {{10}^{14}}\,J}}{{8.192 \times {{10}^{13}}\,J}} \\\ = 7.28 \\\$

The option closest to this answer is option D.

Note: The value of energy of hydrogen is in joule. So, before taking the ratio of energies, remember to convert the value of total energy of uranium that we got in $MeV$ to the corresponding value in joule. $1\,MeV = 1.6 \times {10^{ - 13}}\,J$

Formulas to remember:
The number of atoms $n$ in a given mass can be found out using the formula
$n = \dfrac{m}{M} \times {N_A}$
Where, $m$ is the given mass,$M$ is the molecular mass and ${N_A}$ is the Avogadro number.
Total energy in a given mass is
${E_1} = n \times E$
Where, $E$ is the energy per nucleus.