Calculate the total energy released if 1 kg of 235U undergoes fission? Taking the disintegration energy per event to be Q = 208 MeV. (Answer: 5.32 x 10⁻²⁶ MeV)
Given Data:
Mass of 235U = 1 kg
Q = 208 MeV
Atomic mass of 235U = M = 235 kgK⁻¹mol⁻¹
Avogadro number = NA = 6.023 x 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 NA) in 1 kg given mass 235U So,
n = no. of mole = given mass (m) / Atomic mass (M)
n = 1kg235kgK⁻¹mol⁻¹ = 4.255 x 10⁻³ K mol
N = No. of atom in 235U = n NA
N = 4.255 x 10⁻³ K mol x 6.023 x 10²² atom K⁻¹ mol⁻¹
N = 2.56 x 10²⁴ atoms
Now putting values in equation (1)
E = 2.56 x 10²⁴ atoms x 208 MeV
E = 532.48 x 10²⁴ MeV
E = 5.33 x 10²⁶ MeV -----------------Ans.
************************************
************************************
Shortcut Links For
1. Website for School and College Level Physics 2. Website for School and College Level Mathematics 3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write me in the comment box below for any query and also Share this information with your class-fellows and friends.
1. Website for School and College Level Physics
2. Website for School and College Level Mathematics
3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write me in the comment box below for any query and also Share this information with your class-fellows and friends.
0 Comments
If you have any QUESTIONs or DOUBTS, Please! let me know in the comments box or by WhatsApp 03339719149