Allowed values for the quantum numbers of electrons are as follows:
\(n\) = 1, 2, 3, ...
\(l\) = 0, 1, 2, 3, ..., \(n\) - 1
\(m_{l}\) = 0, \(\pm\)1, \(\pm\)2, \(\pm\)3,..., \(\pm l\)
\(m_{s}\) = \( \pm {1 \over 2} \)
The relationships between n and the shell designations are noted in Table 2.1. relative to the subshells,
\(l\) = 0 corresponds to an \(s\) subshell
\(l\) = 1 corresponds to a \(p\) subshell
\(l\) = 2 corresponds to a \(d\) subshell
\(l\) = 3 corresponds to an \(f\) subshell
For the K shell, the four quantum numbers for each of the two electrons in the 1\(s\) state, in the order of \(nlm_{l}m_{s}\), are 100\(1\over 2\) and 100(-\(1\over 2\)). Write the four quantum numbers for all of the electrons in the L and M shells, and note which correspond to the s, p, and d subshells.
50Cr: 4.34% A = 49.9460 amu
52Cr: 83.79% A = 51.9405 amu
53Cr: 9.50% A = 52.9407 amu
54Cr: 2.37% A = 53.9389 amu
Prove that \(A_{Cr}\) = 51.9963 amu
\(A = f_{1} * A_{1} + f_{2} * A_{2} + f_{3} * A_{3} + f_{4} * A_{4} \)
4.34% * 49.9460 + 83.79% * 51.9405 + 9.50% * 52.9407 + 2.37% * 53.9389 = 51.9963 amu
51.9963 amu = 51.9963 amu \( \implies \) correct