In the fission reaction \(\rm { }_{92}^{235} U +{ }_0^1 n \rightarrow{ }_{56}^{140} Ba +{ }_{36}^{93} Kr +3{ }_0^1 n\) for given masses of 235U (235.0439 amu), 10Ba (139.9106 amu), 93Kr (92.9313 amu) and n (1.00867 amu), and 1 amu (931.494 MeV/e2), the energy released is
172.0 MeV
Fission Reaction Energy Release Calculation
In a nuclear fission reaction, energy is released because the total mass of the products is less than the total mass of the reactants. This difference in mass is called the mass defect ($\Delta m$), which is converted into energy according to Einstein's mass-energy equivalence principle, \(E = \Delta m c^2\).
The given fission reaction is:
\( { }_{92}^{235} U +{ }_0^1 n \rightarrow{ }_{56}^{140} Ba +{ }_{36}^{93} Kr +3{ }_0^1 n \)
We are given the masses of the involved particles:
- Mass of \( { }^{235}U \): 235.0439 amu
- Mass of \( { }^{140}Ba \): 139.9106 amu
- Mass of \( { }^{93}Kr \): 92.9313 amu
- Mass of \( n \): 1.00867 amu
We are also given that 1 amu corresponds to 931.494 MeV/\(c^2\).
Calculating Total Mass of Reactants
The reactants are one \( { }^{235}U \) atom and one neutron (\( n \)).
Total mass of reactants = Mass(\( { }^{235}U \)) + Mass(\( n \))
Total mass of reactants = 235.0439 amu + 1.00867 amu
Total mass of reactants = 236.05257 amu
Calculating Total Mass of Products
The products are one \( { }^{140}Ba \) atom, one \( { }^{93}Kr \) atom, and three neutrons (\( 3n \)).
Total mass of products = Mass(\( { }^{140}Ba \)) + Mass(\( { }^{93}Kr \)) + 3 \(\times\) Mass(\( n \))
Total mass of products = 139.9106 amu + 92.9313 amu + 3 \(\times\) 1.00867 amu
Total mass of products = 139.9106 amu + 92.9313 amu + 3.02601 amu
Total mass of products = 235.86791 amu
Calculating Mass Defect
The mass defect (\(\Delta m\)) is the difference between the total mass of reactants and the total mass of products.
\(\Delta m\) = Total mass of reactants - Total mass of products
\(\Delta m\) = 236.05257 amu - 235.86791 amu
\(\Delta m\) = 0.18466 amu
Calculating Energy Released
The energy released (\(E\)) is calculated using the mass defect and the given energy equivalent of 1 amu.
\(E = \Delta m \times \text{(Energy equivalent of 1 amu)}\)
\(E = 0.18466 \text{ amu} \times 931.494 \text{ MeV/amu}\)
Now, let's calculate the value:
\(E \approx 172.0102 \text{ MeV}\)
Rounding to one decimal place, we get approximately 172.0 MeV.
Summary of Calculation
| Quantity | Value (amu) |
|---|---|
| Mass of reactants | 236.05257 |
| Mass of products | 235.86791 |
| Mass Defect (\(\Delta m\)) | 0.18466 |
Energy Released = \(\Delta m \times 931.494 \text{ MeV/amu}\)
Energy Released = 0.18466 \(\times\) 931.494 MeV \(\approx\) 172.0 MeV
This calculated energy matches one of the given options.
Important Questions from Nuclear Chemistry
For the following nuclear decay series segment,
\(_{90}^{234}{Th}\) → → → \(_{90}^{230}{Th}\)
the overall emitted particles are
α particle is charged ___
Which of the following is used for the production of Nuclear energy?
Which one of the following reactions is the main cause of the energy radiation from the sun
Tritium is an isotope of hydrogen which is radioactive. It decays by _____________.
