Nuclear Physics | NEB Physics | Numerical Problems

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In \(_{Z}X^A\), Z is the number of protons, A is the atomic mass number (i.e., sum of number of protons and number of neutrons) and (A-Z) gives the number of neutrons.
Calculate the binding energy per nucleon of ₂₆Fe⁵⁶. Atomic mass of ₂₆Fe⁵⁶ is 55.9349 u and that of ₁H¹ is 1.00783 u. Mass of ₀n¹ = 1.00867 u and 1 u = 931 MeV.
Solution:
Given,
mass of ₂₆Fe⁵⁶, M = 55.9349 u
mass of ₁H¹ i.e., mass of proton, m_p = 1.00783 u
mass of ₀n¹, m_n = 1.00867 u
1 u = 931 MeV
binding energy per nucleon, \(\Delta E_{ben}\) = ?
The relation to find the binding energy (when the mass is in atomic mass unit (u)) is, \[\Delta E_{be}=\Delta m \times 931\] Let's find \(\Delta m\) which is the mass defect, \[\begin{align*} \Delta m &=\text{theoretical mass of Fe nucleus} - \text{observed mass of Fe nucleus}\\ &=[Zm_p +(A-Z) m_n] - M\\ &=[26 \times 1.00783 + (56 - 26) \times 1.00867] - 55.9349\\ &=0.52878 \hspace{0.1cm}u\\ \end{align*}\] Now, using the relation of binding energy, \[\begin{align*} \Delta E_{be} &=\Delta m \times 931\\ &=0.52878 \times 931\\ \therefore \Delta E_{be}&=492.29 \hspace{0.1cm} \text{MeV}\\ \end{align*}\]
Then, the binding energy per nucleon is, \[\begin{align*} \Delta E_{ben}&=\frac{\Delta E_{be}}{A}\\ &=\frac{492.29}{56}\\ &=8.79 \hspace{0.1cm} \text{MeV}\\ \end{align*}\]
Thus, the required binding energy per nucleon is 8.79 MeV.

A city requires 10⁷ watts of electrical power on the average. If this is to be supplied by a nuclear reactor of efficieny 20 %. Using \(^{92}U_{235}\) as the fuel source, calculate the amount of fuel required per day (Energy released per fission of \(^{92}U_{235}\) = 200 MeV).
Solution:
Given,
power requirement of city, P₀ = 10⁷ W
efficiency, η = 20 %
amount of Uranium fuel required per day = ?
energy released per fission of \(^{92}U_{235}\) = 200 MeV = 200×1.6×10^-13 J = 3.2×10^-11 J

The efficiency of the reactor is given by, \[\eta=\frac{P_o}{P_i} \times 100\%\] where P_i is the power to be generated by the Uranium fuel to provide the output power equal to 10⁷ Watt.
So, \[\begin{align*} \eta&=\frac{P_o}{P_i} \times 100\%\\ 20&=\frac{10^7}{P_i} \times 100\%\\ P_i&=5\times 10^7 \hspace{0.1cm} \text{W}\\ \end{align*}\]
Energy to be released from \(^{92}U_{235}\) is, \[\begin{align*} &=P_i \times t\\ &=5\times 10^7 \times 86400 \\ &=4.32 \times 10^{12} \hspace{0.1cm} \text{J}\\ \end{align*}\]
Energy released in fission of 1 atom is \(3.2 \times 10^{-11}\) J.

To produce 1 J of energy, \(\frac{1}{3.2 \times 10^{-11}}\) atoms undergo fission.

To produce \(4.32 \times 10^{12}\) J of energy, \(\frac{4.32 \times 10^{12}}{3.2 \times 10^{-11}}\) = \(1.35\times 10^{23}\) atoms undergo fission.

Also,

235 g of \(^{92}U_{235}\) contains 6.023×10²³ atoms of \(^{92}U_{235}\).

In 1 atom of \(^{92}U_{235}\), \(\frac{235}{6.023 \times 10^{23}}\) g of \(^{92}U_{235}\) is contained.

In \(1.35 \times 10^{23}\) atoms of \(^{92}U_{235}\), \(\frac{235 \times 1.35 \times 10^{23}}{6.023\times 10^{23}}\) g = 52.67 g of \(^{92}U_{235}\) is contained.

Thus, the amount of fuel required is 52.67 g or 0.0527 kg .

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