The energy released when $\frac{7}{17.13} \text{ kg}$ of $^7_3\text{Li}$ is converted into $^4_2\text{He}$ by proton bombardment is $\alpha \times 10^{32} \text{ eV}$. The value of $\alpha$ is . . . . . . . (Nearest integer) (Mass of $^7_3\text{Li} = 7.0183 \text{ u}$, mass of $^4_2\text{He} = 4.004 \text{ u}$, mass of proton $= 1.008 \text{ u}$, $1 \text{ u} = 931 \text{ MeV/c}^2$, and Avogadro number $N_A = 6.0 \times 10^{23} \text{ mol}^{-1}$)

  • A
    $5$
  • B
    $6$
  • C
    $8$
  • D
    $10$

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Similar Questions

Fission of nuclei is possible because the binding energy per nucleon in them

$A$ nucleus with mass number $240$ breaks into two fragments each of mass number $120$. The binding energy per nucleon of the unfragmented nucleus is $7.6 \, MeV$,while that of the fragments is $8.5 \, MeV$. The total gain in the binding energy in the process is (in $MeV$):

The explosive in a Hydrogen bomb is a mixture of ${ }_1 H^2, { }_1 H^3$ and ${ }_3 Li^6$ in some condensed form. The chain reaction is given by:
${ }_3 Li^6 + { }_0 n^1 \rightarrow { }_2 He^4 + { }_1 H^3$
${ }_1 H^2 + { }_1 H^3 \rightarrow { }_2 He^4 + { }_0 n^1$
During the explosion,the energy released is approximately:
[Given: $M(Li^6) = 6.01690 \ amu, M({ }_1 H^2) = 2.01471 \ amu, M({ }_2 He^4) = 4.00388 \ amu$,and $1 \ amu = 931.5 \ MeV$] (in $MeV$)

Energy released when two deuterons $\left({ }_1 H ^2\right)$ fuse to form a helium nucleus $\left({ }_2 He ^4\right)$ is $:$
(Given $:$ Binding energy per nucleon of ${ }_1 H ^2=1.1 \ \text{MeV}$ and binding energy per nucleon of ${ }_2 He ^4=7.0 \ \text{MeV}$) (in $\text{MeV}$)

In nuclear fission,the fission reaction proceeds with a projectile. Which of the following suits the best?

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