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goldfiish [28.3K]
3 years ago
5

A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given t

hat the mass of the deuterium atom is 2.014102 u and the masses of a hydrogen atom and a neutron are 1.007825 u and 1.008665 u, respectively.
a. 2 GeV.
b. 2 keV.
c. 2 MeV.
d. 2 eV.
Physics
1 answer:
Mademuasel [1]3 years ago
8 0

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

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