Q28.A common example of alpha decay is 23892U →23490 Th +2 He4 + Q Given: 23892U = 238. 05060 u 23490Th = 234. 04360 u 42He = 4. 00260 u and 1u = 931. 5 MeVc2 The energy released (Q) during the alpha decay of 23892U is _____ MeV.
What This Question Tests
This question calculates the energy released in an alpha decay by determining the mass defect and converting it to energy using Einstein's mass-energy equivalence.
Concepts Tested
Formulas Used
Q = (Σm_reactants - Σm_products)c^2
1u = 931.5 MeV/c^2
📚 NCERT Sections This Tests
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
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).
📋 Question Details
- Chapter
- Nuclei
- Topic
- Alpha decay, Mass defect, Energy released (Q-value)
- Year
- 2023
- Shift
- 12 Apr Shift 1
- Q Number
- Q28
- Type
- Numerical
- NCERT Ref
- Class 12 Physics Ch 13: Nuclei
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