The energy equivalent of 1 g of substance is :
(1) 5.6 1026 MeV
By Einstein's mass–energy relation,
E = mc2 = (1 10-3)(3 108)2 = 9 1013 J
Since 1 MeV = 1.6 10-13 J,
E = 9 10131.6 10-13 MeV 5.6 1026 MeV
Answer: (1)
A star has 100\% helium composition. It starts to convert three 4He into one 12C via triple alpha process as 4He + 4He + 4He 12C + Q. The mass of the star is 2.0 1032 kg and it generates energy at the rate of 5.808 1030 W. The rate of converting these 4He to 12C is n 1042 s-1, where n is . [Take, mass of 4He = 4.0026 u, mass of 12C = 12 u]
The energy released in the fusion of 2 kg of hydrogen deep in the sun is EH and the energy released in the fission of 2 kg of 235U is EU. The ratio EHEU is approximately: (Consider the fusion reaction as 4 11H + 2e- 24He + 2 + 6 + 26.7 MeV, energy released in the fission reaction of 235U is 200 MeV per fission nucleus and NA = 6.023 1023)
Binding energy of a certain nucleus is 18 108 J. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:
If M0 is the mass of isotope 125B, Mp and Mn are the masses of proton and neutron, then nuclear binding energy of isotope is:
In a hypothetical fission reaction 92X236 56Y141 + 36Z92 + 3R The identity of emitted particles (R) is :
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is 10-2 MeV. (Given 1 u = 931 MeV/c2, atomic mass of helium = 4.002603 u)
Which of the following nuclear fragments corresponding to nuclear fission between neutron (01n ) and uranium isotope (92235U ) is correct :
The disintegration energy Q for the nuclear fission of 235U 140Ce + 94Zr + n is MeV. Given atomic masses of 235U : 235.0439 u; 140Ce : 139.9054 u, 94Zr : 93.9063 u; n : 1.0086 u, Value of c2 = 931 MeV/u.
The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is 1000x, where x is .
From the statements given below : Choose the correct answer from the options given below :