The binding energy per nucleon of deuteron ( _ 1 ^ 2 H ) and helium nucleus ( _ 2 ^ 4 He…
Physics · JEE Main · NTA Exams — Atoms and Nuclei
The binding energy per nucleon of deuteron \(\left({ }_{1}^{2} \mathrm{H}\right)\) and helium nucleus \(\left({ }_{2}^{4} \mathrm{He}\right)\) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium, nucleus then the energy released is
- 13.9 MeV
- 26.9 MeV
- 23.6 MeV
- 19.2 MeV
Answer
(C) 23.6 MeV
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- An element has binding energy of 8 eV/nucleon. If it has total binding energy of 128 eV then the number of…
- What will be the relation between wavelength of photon and electron of same energy?
- It is proposed to use the nuclear fusion reaction: _ 1 H ^ 2 + _ 1 H ^ 2 _ 2 He ^ 4 in a nuclear reactor of…
- If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements…
- The activity of a radioactive sample is measured as 9750 counts (minute) –1 at t = 0 and 975 counts (minute)…
- Consider the following reaction _ 1 H ^ 2 + _ 1 H ^ 2 _ 2 He ^ 4 + Q If m ( 1 H 2 ) = 2.0141 u ; m ( 2 He 4 )…
- Radioactive element decays to form a stable nuclide, then the rate of decay of reactant ( dN dt ) will vary…
- The ratio of the energies of the hydrogen atom in the first to the second excited state
More Atoms and Nuclei questions · Browse all practice questions