Question

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

Answer 1

Answer:

binding energy =mass defect

c=2

(m)c

2

=(931)MeV

=3.5*931=2358.5MeV

Explanation:

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Answer 2

Answer:

The binding energy of nucleus is 3269MeV

Explanation:

Given that,

The mass defect in a nucleus is $m_{d}=3.5a.m.u$ and we are required to find its binding energy.

The binding energy of a nucleus is defined as the energy required to separate nucleus into sub atomic particles and it is equivalent to the product of its mass defect and squared velocity of light.This can be mathematically written as

$E_{b}=m_{d}c^2$

Where $c=3\times10^8m/s$

Therefore,

$E_{b}=(3.5\times1.67\times10^{-27}kg)(3\times10^8m/s)^2\\=52.605\times10^{-11}J$

We know that $1eV=1.609\times10^{-19}J$ hence the binding energy is

$\frac{52.605\times10^{-11}}{1.609\times10^{-19}}eV\\=32.69\times10^8eV=3269\times10^6eV\\=3269MeV$

Therefore,

The binding energy of nucleus is 3269MeV

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