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VladimirAG [237]
3 years ago
10

What is the binding energy (in MeV per nucleon) for the ⁶₃Li nucleus?

Physics
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

Explanation:

⁶₃Li will have 3 protons and 3 neutrons .

mass of proton in amu = 1.00727 amu

mass of neutron in amu = 1.00866 amu

mass of lithium nucleus in amu = 6.01512 amu

mass defect = 3 (  1.00727  + 1.00866 ) - 6.01512 amu

= .03267 amu

Binding energy = mass defect in amu x 931 Mev

= 30.41 MeV

binding energy per nucleon

no of nucleon = 3 + 3 = 6

binding energy per nucleon = 30.41 / 6 Mev

= 5.068 MeV .

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hichkok12 [17]
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7 0
3 years ago
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4 0
3 years ago
There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An α particle carries a charge of +2
qaws [65]

Answer:

r = 2.84 \times 10^{-14} m

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

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here kinetic energy of the incident alpha particle is given as

KE = 6.4 \times 10^{-13} J

now we have

6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}

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7 0
3 years ago
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Natali [406]
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11111nata11111 [884]

Answer:

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3 0
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