Nucleus $A$ having $Z=17$ and equal number of protons and neutrons has $12 MeV$ binding energy per nucleonAnother nucleus $B$ of $Z=12$ has total 26 nucleons and $18 MeV$ binding energy per nucleons The difference of binding energy of $B$ and $A$ will be ___$MeV$

Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Answer 1

Step 1: Calculate the Mass Number of Nucleus A

Nucleus A has $Z = 17$ (number of protons). Since it has an equal number of protons and neutrons, the number of neutrons is also 17. The mass number ($A$) is the sum of protons and neutrons:

\[ A = 17 + 17 = 34 \]

Step 2: Calculate the Total Binding Energy of Nucleus A

The binding energy per nucleon for nucleus A is $1.2 \, \text{MeV}$. The total binding energy is the product of the binding energy per nucleon and the mass number:

\[ BE_A = 1.2 \times 34 = 40.8 \, \text{MeV} \]

Step 3: Calculate the Total Binding Energy of Nucleus B

Nucleus B has 26 nucleons, and the binding energy per nucleon is $1.8 \, \text{MeV}$:

\[ BE_B = 1.8 \times 26 = 46.8 \, \text{MeV} \]

Step 4: Calculate the Difference in Binding Energies

The difference in binding energies is:

\[ \Delta BE = BE_B - BE_A = 46.8 - 40.8 = 6 \, \text{MeV} \]

Conclusion: The difference in binding energy is $6 \, \text{MeV}$.

Answer 2