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valina [46]
3 years ago
9

Given the following information: Mass of proton = 1.00728 amu Mass of neutron = 1.00866 amu Mass of electron = 5.486 × 10^-4 amu

Speed of light = 2.9979 × 10^8 m/s Calculate the nuclear binding energy (absolute value) of 3Li^6. which has an atomic mass of 6.015126 amu. J/mol.
Chemistry
1 answer:
lesya692 [45]3 years ago
3 0

<u>Answer:</u> The nuclear binding energy of the given element is 2.938\times 10^{12}J/mol

<u>Explanation:</u>

For the given element _3^6\textrm{Li}

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu

M = mass of nucleus = (n_p\times m_p)+(n_n\times m_n)

M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu

Calculating mass defect of the nucleus:

\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So,  0.032694g/mol=0.032694\times 10^{-3}kg/mol

To calculate the nuclear binding energy, we use Einstein equation, which is:

E=\Delta mc^2

where,

E = Nuclear binding energy = ? J/mol

\Delta m = Mass defect = 0.032694\times 10^{-3}kg/mol

c = Speed of light = 2.9979\times 10^8m/s

Putting values in above equation, we get:

E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol

Hence, the nuclear binding energy of the given element is 2.938\times 10^{12}J/mol

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