$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. The speed of light is $c$. The binding energy per nucleon for the parent nucleus is $E_1$ and that for the daughter nuclei is $E_2$. Then:

  • A
    $E_1 = 2E_2$
  • B
    $E_2 = 2E_1$
  • C
    $E_1 > E_2$
  • D
    $E_2 > E_1$

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Similar Questions

$M_p$ denotes the mass of a proton and $M_n$ 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):

The masses of neutron,proton and deuteron in amu are $1.00893$,$1.00813$ and $2.01473$,respectively. The packing fraction of the deuteron in amu is

From the given data,the amount of energy required to break the nucleus of aluminium ${ }_{13}^{27} {Al}$ is $x \times 10^{-3} {J}$.
Mass of neutron $= 1.00866 \, {u}$
Mass of proton $= 1.00726 \, {u}$
Mass of aluminium nucleus $= 26.98154 \, {u}$
(Assume $1 \, {u}$ corresponds to $1 \, {J}$ of energy for the purpose of this calculation)
(Round off to the nearest integer)

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by $E = \frac{3}{5} \frac{Z(Z-1) e^2}{4 \pi \varepsilon_0 R}$. The measured masses of the neutron,${ }_1^1 H$,${ }_7^{15} N$,and ${ }_8^{15} O$ are $1.008665 \ u$,$1.007825 \ u$,$15.000109 \ u$,and $15.003065 \ u$,respectively. Given that the radii of both the ${ }_7^{15} N$ and ${ }_8^{15} O$ nuclei are the same,$1 \ u = 931.5 \ MeV/c^2$ ($c$ is the speed of light),and $e^2 / (4 \pi \varepsilon_0) = 1.44 \ MeV \ fm$. Assuming that the difference between the binding energies of ${ }_7^{15} N$ and ${ }_8^{15} O$ is purely due to the electrostatic energy,the radius of either of the nuclei is $(1 \ fm = 10^{-15} \ m)$: (in $fm$)

In a nuclear fission process,a nucleus $A$ divides into two nuclei $B$ and $C$. If their binding energies are ${E_a}$,${E_b}$,and ${E_c}$ respectively,which of the following relations is correct?

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