Question

Why are elements over a certain size being multiplied by 0.000549u?

This is from the answer sheet.

mass of reactants = [(235.04393 u – (92 x 0.000549u)] + 1.00867 = 236.00209 u mass of products = [141.92971 u – (54 x 0.000549u)]+[89.90774 u – (38 x 0.000549u)]+ (4 x 1.00867 u) = 235.8216 u mass defect = 236.00209 u – 235.8216 u = 0.1805 u = 0.1805 x 1.66054 x 10-27kg = 2.9973 x 10-28 kg

E = m x c2 = 2.9973 x 10-28 kg x (3 x 108 m s-1)2 = 2.698 x 10-11 J = 2.698 x 10-11 J (÷1.60 x 10-13 J) = 169 MeV

Why are they multiplying the no of protons by 0.000549u?

Answer 1

The reason "Why are they multiplying the no. of protons by `0.000549u`" is actually that, Atomic number of the element is being multiplied by the mass of an electron.

Explanation:

When any atom of an element is disassembled into its constituent particles, the well-separated individual particles, we have the following masses:

`p = 1.007276u, n= 1.008664u, e = 0.000549u`.

We know that mass of an electron is `=9.10938356 × 10^-31 kg`
Also that 1 atomic mass unit `=1.660540 xx 10^-27 kg`
Mass of electron in amu `=(9.10938356 × 10^-31)/(1.660540 xx 10^-27)` is the required value.