Calculate the total energy released if 1 kg of `\ ^{235}U` undergoes fission? Taking the disintegration energy per event to be Q = 208 MeV. (Answer: 5.32 ï½˜ 10⁻²⁶ MeV) 



Given Data:

Mass of `\ ^{235}U`  = 1 kg

Q = 208 MeV

Atomic mass of `\ ^{235}U` = M = 235 kgK⁻¹mol⁻¹

Avogadro number = `\N_A` = 6.023 ï½˜ 10²² atom K⁻¹ mol⁻¹



To Find:

The total energy released = E =?



Solution:


The formula for energy released during fission is

E = No. of atoms in a sample x  disintegration energy per event

E = NQ ------------------(1)


We have to find first the No. of atom (N = n `\N_A`) in 1 kg given mass  `\ ^{235}U` So,


n = no. of mole = given mass (m) / Atomic mass (M)


n = `\frac {1 kg}{235 kg K⁻¹ mol⁻¹}` = 4.255 ï½˜ 10⁻³ K mol


 N = No. of atom in  `\ ^{235}U` = n `\N_A`

 N = 4.255 ï½˜ 10⁻³ K mol ï½˜ 6.023 ï½˜ 10²² atom K⁻¹ mol⁻¹

N = 2.56 ï½˜ 10²⁴ atoms


Now putting values in equation (1)


E = 2.56 ï½˜ 10²⁴ atoms ï½˜ 208 MeV


E = 532.48 ï½˜ 10²⁴ MeV


E = 5.33 ï½˜ 10²⁶ MeV -----------------Ans.




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