What is the binding energy of ${ }_{14}^{29} Si$ whose atomic mass is $28.976495 u$ (in $MeV$)?
Mass of proton $= 1.007276 u$
Mass of neutron $= 1.008664 u$
(Neglect the electron mass) (Assume $1 u = 931.5 MeV$)

  • A
    $237.86$
  • B
    $421.72$
  • C
    $387.21$
  • D
    $116.35$

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Similar Questions

The atomic mass of ${ }_{6} C^{12}$ is $12.000000 \ u$ and that of ${ }_{6} C^{13}$ is $13.003354 \ u$. The required energy to remove a neutron from ${ }_{6} C^{13}$,if the mass of a neutron is $1.008665 \ u$,will be: (in $MeV$)

Which of the following statement$(s)$ is/are true in respect of nuclear binding energy?
$(i)$ The mass energy of a nucleus is larger than the total mass energy of its individual protons and neutrons.
(ii) If a nucleus could be separated into its nucleons, an energy equal to the binding energy would have to be transferred to the particles during the separating process.
(iii) The binding energy is a measure of how well the nucleons in a nucleus are held together.
(iv) The nuclear fission is somehow related to acquiring higher binding energy.

If $M(A, Z)$,$M_p$,and $M_n$ represent the masses of the nucleus ${}_{Z}^{A}X$,proton,and neutron in $u$ units respectively $(1u = 931.5 \, MeV/c^2)$,and $BE$ represents the binding energy in $MeV$,then which of the following relations is correct?

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:
$m(_{20}^{40} Ca) = 39.962591 \; u$
$m(_{20}^{41} Ca) = 40.962278 \; u$
$m(_{13}^{26} Al) = 25.986895 \; u$
$m(_{13}^{27} Al) = 26.981541 \; u$
(Given mass of neutron $m_n = 1.008665 \; u$)

For a nucleus ${ }_Z^A X$ having mass number $A$ and atomic number $Z$:
$A.$ The surface energy per nucleon $(b_s) = a_1 A^{2/3}$
$B.$ The Coulomb contribution to the binding energy $b_c = -a_2 \frac{Z(Z-1)}{A^{4/3}}$
$C.$ The volume energy $b_v = a_3 A$
$D.$ Decrease in the binding energy is proportional to surface area.
$E.$ While estimating the surface energy,it is assumed that each nucleon interacts with $12$ nucleons,($a_1, a_2$ and $a_3$ are constants)
Choose the most appropriate answer from the options given below:

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