The energy equivalent of $0.5\, g$ of a substance is $........\, J$.

  • A
    $0.5 \times 10^{13}$
  • B
    $4.5 \times 10^{16}$
  • C
    $4.5 \times 10^{13}$
  • D
    $1.5 \times 10^{13}$

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Similar Questions

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:

Degenerate electron pressure will not be sufficient to prevent the core collapse of a white dwarf if its mass becomes $n$ times the solar mass. The value of $n$ is:

If a $H_2$ nucleus (deuteron) is completely converted into energy,the energy produced will be around .......... $MeV$.

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)

$A$ 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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