Advertisements
Advertisements
प्रश्न
Suppose, we think of fission of a `""_26^56"Fe"` nucleus into two equal fragments `""_13^28"Al"`. Is the fission energetically possible? Argue by working out Q of the process. Given `"m"(""_26^56 "Fe") = 55.93494 "u"` and `"m"(""_13^28 "Al") = 27.98191 "u"`.
Advertisements
उत्तर
The fission of `""_26^56"Fe" ` can be given as:
\[\ce{^56_13 Fe -> 2 ^28_13 Al}\]
It is given that:
Atomic mass of `"m"(""_26^56 "Fe")` = 55.93494 u
Atomic mass of `m(""_13^28 "Al")` = 27.98191 u
The Q-value of this nuclear reaction is given as:
Q = `["m"(""_26^56 "Fe") - 2"m"(""_13^28 "Al")]"c"^2`
`= [55.93494 - 2 xx 27.98191]"c"^2`
`= (-0.02888 " c"^2) "u"`
But 1 u = 931.5 `"MeV/c"^2`
`therefore "Q" = - 0.02888 xx 931.5 = - 26.902` MeV
The Q-value of the fission is negative. Therefore, fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.
संबंधित प्रश्न
The fission properties of `""_94^239"Pu"` are very similar to those of `""_92^235 "U"`. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure `""_94^239 "Pu"` undergo fission?
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much `""_92^235"U"` did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of `""92^235"U"` and that this nuclide is consumed only by the fission process.
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.
In a typical fission reaction, the nucleus is split into two middle-weight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy? Which one has greater liner momentum?
If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimise the energy?
The mass of a neutral carbon atom in ground state is
A free neutron decays to a proton but a free proton does not decay to a neutron. This is because
Calculate the energy released by 1g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope 235U has an abundance of 0.7% by weight in natural uranium.
A uranium reactor develops thermal energy at a rate of 300 MW. Calculate the amount of 235U being consumed every second. Average released per fission is 200 MeV.
Calculate the Q-value of the fusion reaction 4He + 4He = 8Be. Is such a fusion energetically favourable? Atomic mass of 8Be is 8.0053 u and that of 4He is 4.0026 u.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
Calculate the energy that can be obtained from 1 kg of water through the fusion reaction 2H + 2H → 3H + p. Assume that 1.5 × 10−2% of natural water is heavy water D2O (by number of molecules) and all the deuterium is used for fusion.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
Assuming that about 200 MeV of energy is released per fission of 92U235 nuclei, then the mass of U235 consumed per day in a fission reactor of power 1 megawatt will be approximately ______.
In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose because ____.
If in nuclear fusion process the masses of the fusing nuclei be m1 and m2 and the mass of the resultant nucleus be m3, then:
Power generated by a nuclear reactor is given by P = n E / t. Here n represents:
\[\ce{^290_80X ->[\alpha] Y ->[e+] Z ->[\beta-] P ->[e-] Q}\]
In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are:
