Orbital Angular Momentum Ladder Operator Normalisation Ambiguity

Question

Given $J_-=L_-+S_-$, and correspond to total, orbital, and spin, angular momentum lowering operators, respectively.

Where:

$J_-|j, m_j\rangle=\hbar\sqrt{(j+m_j)(j-m_j+1)}|j,m_j-1\rangle$,

$S_-|s\rangle=\hbar|s-1\rangle$ (where $s\geq\frac{1}{2})$ otherwise; $S_-|-\frac{1}{2}\rangle=\hbar|0\rangle$

What normalisation is applied to the result of the $L_-$ operator?

i.e. If $b$ is some value (that may relate directly to $l$) such that $\hbar b$ corresponds to a normalisation: $L_-|l\rangle=\hbar b|l-1\rangle$, what is $b$?

Context: in attempting to answer the following question, I have found un-normalised states and have attributed this to improper understanding of the $L_-$ operator. Please see question below:

"Hydrogen atoms are prepared such that their electron is in a $d$ orbital, carrying total angular momentum $j=5/2$, with $z$ component $m_j=1/2$. Calculate the proportion of elections that will be found on measurement to have their spin pointing upwards, i.e. such that $m_s=1/2$. (Hint: write down an expression for the $m_j=5/2$ state, and then apply $J_-$ in steps.)"

Answer 1

By a separate mark scheme, it appears that $b=\sqrt{2l}$ such that:

$L_-|l\rangle=\hbar\sqrt{2l}|l-1\rangle$.

Source: a similar question from a paper with newly found model answers:

$L_-|3\rangle=\hbar\sqrt{6}|2\rangle$,

$L_-|1\rangle=\hbar\sqrt{2}|0\rangle$.

Note: There is not an explicitly stated $b$ value in the notes we have received for this module, or the examination model answers, feel free to dispute my findings.