The average energy released per fission for the nucleus of ${}_{92}^{235}U$ is 190 MeV. When all the atoms of 47 g pure ${}_{92}^{235}U$ undergo fission process, the energy released is $α \times 10^{23} MeV$ . The value of $α$ is ______.

(Avogadro Number $= 6 \times 10^{23}$ per mole) [2026]

Question

The average energy released per fission for the nucleus of ${}_{92}^{235}U$ is 190 MeV. When all the atoms of 47 g pure ${}_{92}^{235}U$ undergo fission process, the energy released is $α \times 10^{23} MeV$ . The value of $α$ is ______.

(Avogadro Number $= 6 \times 10^{23}$ per mole) [2026]

Smart Achievers · Accepted Answer

(228)

$Total number of U-235 atoms is$

$47 g = \frac{47}{235} moles = \frac{1}{5} moles$

$α$ $Total energy released = \frac{1}{5} \times 6 \times 10^{23} \times 190 MeV$

$= 228 \times 10^{23} MeV$

Solution

Q.
The average energy released per fission for the nucleus of ${}_{92}^{235}U$ is 190 MeV. When all the atoms of 47 g pure ${}_{92}^{235}U$ undergo fission process, the energy released is $α \times 10^{23} MeV$ . The value of $α$ is ______.

(Avogadro Number $= 6 \times 10^{23}$ per mole) [2026]

Ans.
(228)

$Total number of U-235 atoms is$

$47 g = \frac{47}{235} moles = \frac{1}{5} moles$

$∴$ $Total energy released = \frac{1}{5} \times 6 \times 10^{23} \times 190 MeV$

$= 228 \times 10^{23} MeV$

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Solution

Q. The average energy released per fission for the nucleus of U92235 is 190 MeV. When all the atoms of 47 g pure U92235 undergo fission process, the energy released is α×1023 MeV. The value of α is ______. (Avogadro Number =6×1023 per mole) [2026]

Ans. (228) Total number of U-235 atoms is 47 g=47235 moles=15 moles ∴ Total energy released=15×6×1023×190 MeV =228×1023 MeV

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Q.
The average energy released per fission for the nucleus of ${}_{92}^{235}U$ is 190 MeV. When all the atoms of 47 g pure ${}_{92}^{235}U$ undergo fission process, the energy released is $α \times 10^{23} MeV$ . The value of $α$ is ______.

(Avogadro Number $= 6 \times 10^{23}$ per mole) [2026]

Ans.
(228)

$Total number of U-235 atoms is$

$47 g = \frac{47}{235} moles = \frac{1}{5} moles$

$∴$ $Total energy released = \frac{1}{5} \times 6 \times 10^{23} \times 190 MeV$

$= 228 \times 10^{23} MeV$