If a star converts all the helium $(H e)$ nuclei completely into oxygen $(O)$ nuclei, the energy released per oxygen nucleus is (mass of helium nucleus = $4.0026 a m u$ and mass of oxygen nucleus = $15.9994 a m u$ )

11.99

MeV

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10.24

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10.24

MeV

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- 10.24

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Solution

The correct option is C $10.24$ MeV
The reaction is
$4_{2} H e^{4} ⟶_{8} O^{16} + Q$
$Q = Δ m \times 931 M e V$
$Δ m$ (mass defect) = $4 \times 4.0026 - 15.994 = 0.011 a m u$
$∴$ $Q = 0.011 \times 931 M e V$
$= 10.24 M e V$

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Similar questions

If a star converts all the helium $(H e)$ nuclei completely into oxygen $(O)$ nuclei, the energy released per oxygen nucleus is (mass of helium nucleus = $4.0026 a m u$ and mass of oxygen nucleus = $15.9994 a m u$ )

Q. If a star can convert all of the Helium nuclei, completely into oxygen nuclei, the energy released per oxygen nucleus is,
[Mass of Helium nucleus

= 4.0026 amu

and mass of Oxygen nucleus

= 15.9994 amu

]

Q. If a star can convert all the

H e

nuclei completely into oxygen nuclei, the energy released per oxygen nuclei is [Mass of

H e

nucleus is

4.0026 a m u

and mass of oxygen nucleus is

15.9994 a m u

]

Q. If a star can convert all the helium nuclei completely into oxygen nuclei, then the energy released per oxygen nuclei is

[Mass of He nucleus = 4.0026 amu mass of oxygen nucleus = 15.9994 amu, 1 amu = \frac{931.5}{c^{2}} MeV]

Q. If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is [Mass of He nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu]. [IIT-JEE-2005]

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If a star converts all the helium (He) nuclei completely into oxygen (O) nuclei, the energy released per oxygen nucleus is (mass of helium nucleus = 4.0026 amu and mass of oxygen nucleus = 15.9994 amu)

The correct option is C 10.24 MeVThe reaction is 4 2He4⟶ 8O16+Q Q=Δm×931 MeV Δm (mass defect) = 4×4.0026−15.994=0.011 amu ∴ Q=0.011×931 MeV =10.24 MeV

If a star converts all the helium $(H e)$ nuclei completely into oxygen $(O)$ nuclei, the energy released per oxygen nucleus is (mass of helium nucleus = $4.0026 a m u$ and mass of oxygen nucleus = $15.9994 a m u$ )

The correct option is C $10.24$ MeV
The reaction is
$4_{2} H e^{4} ⟶_{8} O^{16} + Q$
$Q = Δ m \times 931 M e V$
$Δ m$ (mass defect) = $4 \times 4.0026 - 15.994 = 0.011 a m u$
$∴$ $Q = 0.011 \times 931 M e V$
$= 10.24 M e V$