Question

Instead of giving atomic masses for nuclides as in Appendix B, some Tables give the mass excess, ?, defined as ?=M?A, where A is the atomic number and M is the mass in u.

A)Determine the mass excess, in u, for 42He; 126C; 10747Ag; 23592U.

b)Determine the mass excess, in MeV/c2, for 42He; 126C; 10747Ag; 23592U.

Answer 1

We know that the atomic masses for the given elements in atomic mass unit is

1) He : 4.002602 u

2) C : 12.0107 u

3) Ag : 107.8682 u

4) U : 235.04393

The mass excess for various elements will be given by

?m(delta m) = M - A

So the mass excess for various elements will be in atomic mass units (u) will be

1) for He ?m = 4.002602 - 4 = 0.002602 u

2) for C ?m = 12.0107 - 12 = 0.0107 u

3) for Ag ?m = 107.8682 - 107 = 0.8682 u

4) for U ?m = 235.04393 - 235 = 0.04393 u

Now the same mass excess in MeV/c^2

We know that 1 u = 931.5 MeV/c^2

So

1) for He

?m = 0.002602 u = 0.002602 x 931.5 = 2.424 MeV/c^2

2) for C ?m = 0.0107 u = 0.0107 x 931.5 = 9.967 MeV/c^2

3) for Ag ?m = 0.8682 u = 0.8682 x 931.5 = 808.729 MeV/c^2

4) for U ?m = 0.04393 u = 0.04393 x 931.5 = 40.921 MeV/c^2

P.S. : wherever it may show question mark there is delta. Sometimes the browser is not able to read the special characters. So wherever it showing question mark (if it is anywhere) there is the delta over their for mass excess.