When _{92}U^{235} undergoes fission,0.1% of i | vedclass

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(D) The mass defect $\Delta m$ is $0.1\%$ of the initial mass $m = 1 \, kg$.$\Delta m = \frac{0.1}{100} 	imes 1 \, kg = 10^{-3} \, kg$.Using Einstein

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$9 	imes 10^{13} \, J$

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(D) The mass defect $\Delta m$ is $0.1\%$ of the initial mass $m = 1 \, kg$.$\Delta m = \frac{0.1}{100} 	imes 1 \, kg = 10^{-3} \, kg$.Using Einstein's mass-energy equivalence formula,$E = \Delta m c^2$,where $c = 3 	imes 10^8 \, m/s$ is the speed of light.$E = 10^{-3} \, kg 	imes (3 	imes 10^8 \, m/s)^2$.$E = 10^{-3} 	imes 9 	imes 10^{16} \, J$.$E = 9 	imes 10^{13} \, J$.

When $_{92}U^{235}$ undergoes fission,$0.1\%$ of its original mass is converted into energy. How much energy is released if $1\,kg$ of $_{92}U^{235}$ undergoes fission?

Similar Questions

Assertion : Energy is released in nuclear fission.
Reason : Total binding energy of the fission fragments is larger than the total binding energy of the parent nucleus.

Explain the phenomenon of nuclear fission.

In an atomic reactor,the kinetic energy of fast-moving neutrons can be reduced by colliding them with:

$A$ nucleus of uranium decays at rest into nuclei of thorium and helium. Then:

Nuclear fusion is common to the pair:

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