Calculate the mass defect of the boron nucleus 11 5b. the mass of neutral 11 5b is equal to 11.009305 atomic mass units.

The mass defect of the Boron : Δm = 0.081815 amu

Atoms are composed of 3 types of basic particles namely protons, electrons, and neutrons

The mass of this basic particle is expressed in atomic mass units (amu)

The mass of atoms is always slightly less than the sum of the masses of the individual atomic forming particles namely neutrons, protons, and electrons. The difference between the mass of the atom and the sum of the masses of atomic subparticles is called the mass defect (Δm).

Can be formulated:

Δm = [Z (mp + me) + (A-Z) mn] - m atom

where:

Δm = mass defect (amu)

mp = mass of a proton (1,007277 amu)

mn = mass of a neutron (1.008665 amu)

me = mass of an electron (0.000548597 amu)

matom = mass of nuclide (amu)

Z = atomic number

A = mass number

Atomic number (Z) = number of protons = number of electrons

Mass Number (A) is the sum of protons and neutrons

Atomic Number (Z) = Mass Number (A) - Number of Neutrons

For Boron elements (₁₁⁵B) which have atomic number 5 and mass number 11, then:

number of protons = 5

number of electrons = 5

number of neutrons = mass number - atomic number = 11 - 5 = 6

then:

Δm = [5 (1.007277 + 0.000548597) + (6) 1.008665] - 11.009305

Δm = [5 (1.007826) + (6) 1.008665] - 11.009305

Δm = (5.03913 + 6.05199) - 11.009305

Δm = 11.09112 - 11.009305

Δm = 0.081815 amu

element symbol

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the chemical symbols

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The subatomic particle that has the least mass  

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Keywords: mass number, atomic mass, amu, mass defect, atomic number

the answer is either b or c