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Calculate the total binding energy, and the binding energy per nucleon, for 73Li. The masses of the atoms of 73Li and 11H are 7.016004 u and 1.007825 u, respectively. The mass of a neutron is 1.008665 u.

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Answer:

5.60 Mev

Explanation:

Binding Energy is termed as the total energy required for the breakdown of a nucleus to its component nucleons

From the given question;

There are 3H and 4n,

Therefore;

since the masses of the atoms are 7.016004 u and 1.007825 u,

Then

3(1.007825) + 4(1.008665) = 7.058135 u

( 7.058135 - 7.016004) × (931.5) = 39.24 MeV total biding energy.

The energy per nucleon is  therefore calculate as:

[tex]\frac {39.24 \ \ MeV }{7}[/tex]

=5.60 Mev

Thus, the total binding energy is = 5.60 Mev

Lanuel

a. The total binding energy for [tex]^7_3Li[/tex] is equal to 39.25 MeV.

b. The binding energy per nucleon for [tex]^7_3Li[/tex] is equal to 5.61 MeV/nucleon.

Given the following data:

  • Mass of lithium (Li) atoms = 7.016004 u.
  • Mass of hydrogen (H) atoms = 1.007825 u.
  • Mass of a neutron = 1.008665 u.

To calculate the total binding energy, and the binding energy per nucleon, for [tex]^7_3Li[/tex]:

So, there are three hydrogen atoms (3H) and four neutrons (4n).

Substituting the given parameters into the formula, we have;

[tex]3(1.007825) + 4(1.008665) = 7.058135 \;u[/tex]

Note: 1 atomic mass unit (a.m.u) is equal to 931.5 MeV.

For the total binding energy:

[tex]T_{BE} =( 7.058135 -7.016004) \times 931.5\\\\T_{BE} =0.042131 \times 931.5[/tex]

Total binding energy = 39.25 MeV.

For the binding energy per nucleon:

[tex]B_E = \frac{39.25}{7}[/tex]

Binding energy per nucleon = 5.61 MeV/nucleon.

Read more on binding energy here: https://brainly.com/question/3405451

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