Advertisements
Advertisements
प्रश्न
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei `""_20^41"Ca"` and `""_13^27 "Al"` from the following data:
`"m"(""_20^40"Ca")` = 39.962591 u
`"m"(""_20^41"Ca")` = 40.962278 u
`"m"(""_13^26"Al")` = 25.986895 u
`"m"(""_13^27"Al")` = 26.981541 u
Advertisements
उत्तर
For `""_20^41"Ca":` Separation energy = 8.363007 MeV
For `""_13^27"Al":` Separation energy = 13.059 MeV
A neutron `(""_0"n"^1)` is removed from a `""_20^41 "Ca"` nucleus. The corresponding nuclear reaction can be written as:
`""_20^41"Ca" -> ""_20^40"Ca" + _0^1"n"`
It is given that:
Mass `"m"(""_20^40 "Ca")`= 39.962591 u
Mass `"m"(""_20^41 "Ca")` = 40.962278 u
Mass m(`""_0"n"^1`) = 1.008665 u
The mass defect of this reaction is given as:
Δ m = `"m"(""_20^40"Ca") + (""_0^1"n") - "m"(""_20^41 "Ca")`
`= 39.962591 + 1.008665 - 40.962278 = 0.008978 "u"`
But 1 u = 931.5 MeV/c2
∴ Δ m = 0.008978 × 931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
`"E" = triangle"mc"^2`
= 0.008978 xx 931.5 = 8.363007 MeV
For `""_13^27 "Al"` the neutron removal reaction can be written as:
\[\ce{^27_13Al -> ^26_13Al + ^1_0n}\]
it us given that.
Mass `"m"(""_13^27 "Al")` = 26.981541 u
Mass `"m"(""_13^26 "Al")` = 25.986895 u
The mass defect of this reaction is given as:
`triangle"m" = "m"(""13^26 "Al") + "m"(""_0^1 "n") - "m"(""_13^27 "Al")`
= 25.986895 + 1.008665 - 26.981541
= 0.014019 u
`= 0.014019 xx 931.5 " MeV/c"^2`
Hence, the energy required for neutron removal is calculated as:
`E = triangle"mc"^2`
= 0.014019 x 931.5 = 13.059 MeV
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
What is meant by the terms half-life of a radioactive substance and binding energy of a nucleus?
Define the terms (i) half-life (T1/2) and (ii) average life (τ). Find out their relationships with the decay constant (λ).
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.)
Calculate mass defect and binding energy per nucleon of `"_10^20 Ne`, given
Mass of `"_10^20 Ne= 19.992397` u
Mass of `"_0^1H = 1.007825` u
Mass of `"_0^1n = 1.008665` u
Sketch a graph showing the variation of binding energy per nucleon of a nucleus with its mass number.
The figure shows the plot of binding energy (BE) per nucleon as a function of mass number A. The letters A, B, C, D, and E represent the positions of typical nuclei on the curve. Point out, giving reasons, the two processes (in terms of A, B, C, D, and E ), one of which can occur due to nuclear fission and the other due to nuclear fusion.
Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′: F = `1/(4πε_0) e^('2)/r` estimate the value of (e’/e) given that the binding energy of a deuteron is 2.2 MeV.
Explain the release of energy in nuclear fission and fusion on the basis of binding energy per nucleon curve.
What is meant by “binding energy per nucleon” of a nucleus?
Find the binding energy per nucleon of 235U based on the information given below.
| Mass(u) | |
| mass of neutral `""_92^235"U"` | 235.0439 |
| mass of a proton | 1.0073 |
| mass of a neutron | 1.0087 |
The binding energy per nucleon of 37Li and 24He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction 3Li7 +1H1 → 2He4 + 2He4 + Q the value of energy Q released is:
Mp denotes the mass of a proton and Mn that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light):
Binding energy per nucleon is maximum for:
