When four hydrogen nuclei combine to form a helium nucleus at the core of our Sun, about $0.01643$ a.m.u. of mass is lost, The approximate amount of energy released in the process, is:

15.30 MeV

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30.60 MeV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16.43 MeV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

32.86 MeV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $15.30 MeV$
Given:
Loss of mass, $Δ m = 0.01643$ a.m.u.

We know that loss of 1 a.m.u. results in an energy of about 931 MeV.

$⟹ 0.01643$ a.m.u. would result in an energy equal to $0.01643 \times 931 = 15.296$ MeV
$\approx 15.3 MeV$

Suggest Corrections

Similar questions

Q. When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass in

0.0625

amu. How much energy is released?

Q. When four hydrogen nuclei fuse together to form a helium nucleus, then in this process

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. The masses of neutron and proton are

1.0087 a . m . u .

and

1.0073 a . m . u .

respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass

4.0015 a . m . u

, then the binding energy fo the helium nucleus will be: (

1 a . m . u . = 931 M e V

)

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 = $15.9994 a m u$ )

Join BYJU'S Learning Program

When four hydrogen nuclei combine to form a helium nucleus at the core of our Sun, about 0.01643 a.m.u. of mass is lost, The approximate amount of energy released in the process, is:

The correct option is A 15.30 MeVGiven: Loss of mass, Δm=0.01643 a.m.u. We know that loss of 1 a.m.u. results in an energy of about 931 MeV. ⟹0.01643 a.m.u. would result in an energy equal to 0.01643×931=15.296 MeV ≈15.3 MeV

When four hydrogen nuclei combine to form a helium nucleus at the core of our Sun, about $0.01643$ a.m.u. of mass is lost, The approximate amount of energy released in the process, is:

The correct option is A $15.30 MeV$
Given:
Loss of mass, $Δ m = 0.01643$ a.m.u.

We know that loss of 1 a.m.u. results in an energy of about 931 MeV.

$⟹ 0.01643$ a.m.u. would result in an energy equal to $0.01643 \times 931 = 15.296$ MeV
$\approx 15.3 MeV$