Q28.The energy released per fission of nucleus of 240X is 200 MeV . The energy released if all the atoms in 120 g of pure 240X undergo fission is _____ ×1025 MeV . (Given NA = 6 × 1023 )
What This Question Tests
This question requires calculating the total energy released from the fission of a given mass of a nucleus, combining concepts of Avogadro's number and energy per fission.
Concepts Tested
Formulas Used
Number of atoms = (mass/molar mass) * NA
Total energy = Number of atoms * energy per fission
📚 NCERT Sections This Tests
13.2 — Obtain The Binding Energy Of The Nuclei 5626Fe And 20983 Bi In Units Of
Physics Class 12 · Chapter 13
13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u
13.3 — A Given Coin Has A Mass Of 3.0 G. Calculate The Nuclear Energy That
Physics Class 12 · Chapter 13
13.3 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u).
13.5 — The Q Value Of A Nuclear Reaction A + B ® C + D Is Defined By
Physics Class 12 · Chapter 13
13.5 The Q value of a nuclear reaction A + b ® C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) 11 H+13 H →12 H+12 H (ii) 126 C+126 C →1020 Ne+ 24 He Atomic masses are given to be m ( 12 H ) = 2.014102 u m ( 13 H) = 3.016049 u m ( 126 C ) = 12.000000 u m ( 1020 Ne ) = 19.992439 u
📋 Question Details
- Chapter
- Nuclei
- Topic
- Nuclear Fission Energy
- Year
- 2023
- Shift
- 24 Jan Shift 2
- Q Number
- Q28
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 13: Nuclei
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