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प्रश्न
Which property of nuclear force explains the constancy of binding energy per nucleon `((BE)/A)` for nuclei in the range 20< A < 170 ?
उत्तर
(1) Nuclear force is short range force
(2) Most of the nuclear reside inside the nucleus and not on surface
(3) Density of nucleus is independent of mass number
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संबंधित प्रश्न
Obtain the binding energy (in MeV) of a nitrogen nucleus `(""_7^14"N")`, given `"m"(""_7^14"N")` = 14.00307 u.
Consider the fission of `""_92^238"U"` by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are `""_58^140"Ce"` and `""_44^99"Ru"`. Calculate Q for this fission process. The relevant atomic and particle masses are
`"m"(""_92^238"U")` = 238.05079 u
`"m"(""_58^140"Ce")` = 139.90543 u
`"m"(""_44^99"Ru")` = 98.90594 u
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How much energy is released in the following reaction : 7Li + p → α + α.
Atomic mass of 7Li = 7.0160 u and that of 4He = 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.)
Answer the following question.
Draw the curve showing the variation of binding energy per nucleon with the mass number of nuclei. Using it explains the fusion of nuclei lying on the ascending part and fission of nuclei lying on the descending part of this curve.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
Calculate the binding energy of an alpha particle in MeV. Given
mass of a proton = 1.007825 u
mass of a neutron = 1.008665 u
mass of He nucleus = 4.002800 u
1u = 931 MeV/c2
State the significance of binding energy per nucleon.
What is meant by “binding energy per nucleon” of a nucleus?
