(B) The initial binding energy of the nucleus is calculated as: $E_{i} = 242 \times 7.6\,MeV = 1839.2\,MeV$.The final state consists of two fragments,

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(B) The initial binding energy of the nucleus is calculated as: $E_{i} = 242 \times 7.6\,MeV = 1839.2\,MeV$.The final state consists of two fragments,each with a mass number of $121$. The binding energy per nucleon for each fragment is $8.1\,MeV$.The total final binding energy is: $E_{f} = (121 \times 8.1\,MeV) + (121 \times 8.1\,MeV) = 242 \times 8.1\,MeV = 1960.2\,MeV$.The total gain in binding energy is the difference between the final and initial binding energies:$\Delta E = E_{f} - E_{i} = 242 \times (8.1 - 7.6)\,MeV$.$\Delta E = 242 \times 0.5\,MeV = 121\,MeV$.

Class 12 Physics Nuclei Mass-Energy, Nuclear Binding Energy, Nuclear Stability

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$A$ nucleus with mass number $242$ and binding energy per nucleon as $7.6\,MeV$ breaks into two fragments,each with mass number $121$. If each fragment nucleus has a binding energy per nucleon of $8.1\,MeV$,the total gain in binding energy is $........MeV$.

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A
$120$
B
$121$
C
$122$
D
$159$

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Nuclei

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Mass-Energy, Nuclear Binding Energy, Nuclear Stability

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If $M(A, Z)$,$M_p$,and $M_n$ denote the masses of the nucleus ${}_Z^AX$,proton,and neutron respectively in units of $u$ $(1u = 931.5 \text{ MeV}/c^2)$,and $BE$ represents its binding energy in $\text{MeV}$,then:

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The atomic mass of ${ }_7 N ^{15}$ is $15.000108 \text{ a.m.u.}$ and that of ${ }_8 O ^{16}$ is $15.994915 \text{ a.m.u.}$ If the mass of a proton is $1.007825 \text{ a.m.u.}$,then the minimum energy required to remove the least tightly bound proton is ......... $MeV$.

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Find the energy equivalent of one atomic mass unit,first in $Joules$ and then in $MeV$. Using this,express the mass defect of $_{8}^{16} O$ in $MeV / c^{2}$.

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$A$ nucleus of mass $ 20 u $ emits a $ \gamma $ photon of energy $ 6 MeV $. If the emission is assumed to occur when the nucleus is free and at rest, then the nucleus will have a kinetic energy nearest to (take $ 1 u = 1.6 \times 10^{-27} kg $): (in $keV$)

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The binding energy per nucleon for $C^{12}$ is $7.68 \text{ MeV}$ and that for $C^{13}$ is $7.5 \text{ MeV}$. The energy required to remove a neutron from $C^{13}$ is ......... $\text{MeV}$.

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$A$ nucleus with mass number $242$ and binding energy per nucleon as $7.6\,MeV$ breaks into two fragments,each with mass number $121$. If each fragment nucleus has a binding energy per nucleon of $8.1\,MeV$,the total gain in binding energy is $........MeV$.

Similar Questions

If $M(A, Z)$,$M_p$,and $M_n$ denote the masses of the nucleus ${}_Z^AX$,proton,and neutron respectively in units of $u$ $(1u = 931.5 \text{ MeV}/c^2)$,and $BE$ represents its binding energy in $\text{MeV}$,then:

The atomic mass of ${ }_7 N ^{15}$ is $15.000108 \text{ a.m.u.}$ and that of ${ }_8 O ^{16}$ is $15.994915 \text{ a.m.u.}$ If the mass of a proton is $1.007825 \text{ a.m.u.}$,then the minimum energy required to remove the least tightly bound proton is ......... $MeV$.

Find the energy equivalent of one atomic mass unit,first in $Joules$ and then in $MeV$. Using this,express the mass defect of $_{8}^{16} O$ in $MeV / c^{2}$.

$A$ nucleus of mass $ 20 u $ emits a $ \gamma $ photon of energy $ 6 MeV $. If the emission is assumed to occur when the nucleus is free and at rest, then the nucleus will have a kinetic energy nearest to (take $ 1 u = 1.6 \times 10^{-27} kg $): (in $keV$)

The binding energy per nucleon for $C^{12}$ is $7.68 \text{ MeV}$ and that for $C^{13}$ is $7.5 \text{ MeV}$. The energy required to remove a neutron from $C^{13}$ is ......... $\text{MeV}$.

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