The mass defect in a nucleus is $3.5$ amu. Then the binding energy of the nucleus is.

3258.5

MeV

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325.85

MeV

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32.58

MeV

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3.258

MeV

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Solution

The correct option is A $3258.5$ MeV
A
Binding energy =(mass defect) $c^{2} = (Δ m) c^{2}$
$= Δ (931) M e V$
$= 3.5 \times 931 = 3258.5 M e V$

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Similar questions

Q. The mass defect for the nucleus of Helium is

0.0303 a m u

. The binding energy per nucleon in

M e V

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]

If the binding energy of deuterium is 2.23 MeV, then the mass defect (in amu) is?

Q. The binding energy of a deuterium nucleus is about

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Q. The mass of the nucleus

_{7} N^{14}

14.00000 a m u

. Calculate the binding energy per nucleon of

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, mass of neutron

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and

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The mass defect in a nucleus is 3.5 amu. Then the binding energy of the nucleus is.

The correct option is A 3258.5 MeVABinding energy =(mass defect) c2=(Δm)c2=Δ(931)MeV=3.5×931=3258.5MeV

The mass defect in a nucleus is $3.5$ amu. Then the binding energy of the nucleus is.

The correct option is A $3258.5$ MeV
A
Binding energy =(mass defect) $c^{2} = (Δ m) c^{2}$
$= Δ (931) M e V$
$= 3.5 \times 931 = 3258.5 M e V$