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, $\mathrm{\Delta }{E}_{ben}$ = ?
The relation to find the binding energy (when the mass is in atomic mass unit (u)) is, $$\mathrm{\Delta }{E}_{be}=\mathrm{\Delta }m\times 931$$ Let's find $\mathrm{\Delta }m$ which is the mass defect, $$\begin{array}{rl}\mathrm{\Delta }m& =\text{theoretical mass of Fe nucleus}-\text{observed mass of Fe nucleus}\\ & =[Z{m}_p+(A-Z)m_n]-M\\ & =[26\times 1.00783+(56-26)\times 1.00867]-55.9349\\ & =0.52878u\end{array}$$ Now, using the relation of binding energy, $$\begin{array}{rl}\mathrm{\Delta }{E}_{be}& =\mathrm{\Delta }m\times 931\\ & =0.52878\times 931\\ \therefore \mathrm{\Delta }{E}_{be}& =492.29\text{MeV}\end{array}$$
Then, the binding energy per nucleon is, $$\begin{array}{rl}\mathrm{\Delta }{E}_{ben}& =\frac{\mathrm{\Delta }{E}_{be}}{A}\\ & =\frac{492.29}{56}\\ & =8.79\text{MeV}\end{array}$$
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, $$\eta =\frac{P_o}{P_i}\times 100\mathrm{\%}$$ where P_i is the power to be generated by the Uranium fuel to provide the output power equal to 10⁷ Watt.
So, $$\begin{array}{rl}\eta & =\frac{P_o}{P_i}\times 100\mathrm{\%}\\ 20& =\frac{{10}^7}{P_i}\times 100\mathrm{\%}\\ P_i& =5\times {10}^7\text{W}\end{array}$$
Energy to be released from ${}^{92}{U}_{235}$ is, $$\begin{array}{rl}& =P_i\times t\\ & =5\times {10}^7\times 86400\\ & =4.32\times {10}^{12}\text{J}\end{array}$$
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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