What will be the energy released in joule, in the process of fission by $1 \text{ mg}$ of ${ }_{92}^{240} U$? Assume energy release per fission is $200 \text{ MeV}$. [Use Avogadro's number as $6 \times 10^{23}$ and $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$]

  • A
    $6.2 \times 10^7 \text{ J}$
  • B
    $7.0 \times 10^7 \text{ J}$
  • C
    $8.0 \times 10^7 \text{ J}$
  • D
    $8.2 \times 10^7 \text{ J}$

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Similar Questions

The $Q$ value of a nuclear reaction $A+b \rightarrow C+d$ is defined by $Q=\left[m_{A}+m_{b}-m_{C}-m_{d}\right] c^{2}$ where the masses refer to the respective nuclei. Determine from the given data the $Q$ value of the following reactions and state whether the reactions are exothermic or endothermic.
$(i) \;_{1}^{1} H+_{1}^{3} H \rightarrow_{1}^{2} H+_{1}^{2} H$
$(ii)\;_{6}^{12} C+_{6}^{12} C \rightarrow_{10}^{20} N e+_{2}^{4} H e$
Atomic masses are given to be:
$m(_{1}^{1}H) = 1.007825 \; u$
$m(_{1}^{2}H) = 2.014102 \; u$
$m(_{1}^{3}H) = 3.016049 \; u$
$m(_{6}^{12}C) = 12.000000 \; u$
$m(_{10}^{20}Ne) = 19.992439 \; u$
$m(_{2}^{4}He) = 4.002603 \; u$

Under certain circumstances,a nucleus can decay by emitting a particle more massive than an $\alpha$-particle. Consider the following decay processes:
$_{88}^{223} Ra \rightarrow _{82}^{209} Pb + _{6}^{14} C$
$_{88}^{223} Ra \rightarrow _{86}^{219} Rn + _{2}^{4} He$
Calculate the $Q$-values for these decays and determine that both are energetically allowed.

If all the helium in a star is converted into oxygen in its core,the energy released per oxygen nucleus is...........$MeV$?
[Mass of $He = 4.0026 \, a.m.u.$,Mass of $O = 15.9994 \, a.m.u.$]

$A$ nuclear reactor generates power by fission of ${}_{92}U^{235}$ into two equal fragments of ${}_{46}Pd^{116}$ with the emission of two $\gamma$-rays of $5.2 \text{ MeV}$ each and three neutrons. The average binding energies per nucleon of ${}_{92}U^{235}$ and ${}_{46}Pd^{116}$ are $7.2 \text{ MeV}$ and $8.2 \text{ MeV}$ respectively. The usable energy released per fission event is ......... $\text{MeV}$.

In the following nuclear reaction, $x$ is: ${ }_{13} Al^{27} + { }_2 He^4 \rightarrow { }_0 n^1 + X$

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