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Binding Energy

The masses of...

Question

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]

28.4 MeV

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20.8 MeV

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27.3 MeV

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14.2 MeV

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Solution

The correct option is A 28.4 MeV
Helium nucleus consist of two neutrons and two protons.
So binding energy $E = Δ m \times 931 M e V$
$\Rightarrow E = (2 \times m_{p} + 2 m_{n} - M) \times 931 M e V = (2 \times 1.0073 + 2 \times 1.0087 - 4.0015) \times 931 = 28.4 M e V$

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Similar questions

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV]

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

)

Q. The binding energy of an

α

-particle is
(Given that mass of proton

=

1.0073

amu, mass of neutron

=

1.0087

amu, and mass of

α

-particle

=

4.0015

amu)

Q. Find the binding energy of an

α -

particle from the following data.
Mass of the helium nucleus

= 4.001265 a m u

Mass of proton

= 1.007277 a m u

Mass of neutron

= 1.00866 a m u

(take

1 a m u = 931.4813 M e V)

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The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]

The correct option is A 28.4 MeV Helium nucleus consist of two neutrons and two protons. So binding energy E=Δm×931MeV ⇒ E=(2×mp+2mn−M)×931MeV=(2×1.0073+2×1.0087−4.0015)×931=28.4MeV

The correct option is A 28.4 MeV
Helium nucleus consist of two neutrons and two protons.
So binding energy $E = Δ m \times 931 M e V$
$\Rightarrow E = (2 \times m_{p} + 2 m_{n} - M) \times 931 M e V = (2 \times 1.0073 + 2 \times 1.0087 - 4.0015) \times 931 = 28.4 M e V$