The number of electrons which can have the quantum numbers $n = 3$ , $l = anything$ , $m_{l} = 1$ and $m_{s} = anything$ is:

Open in App

Solution

For $n = 3$ $\Rightarrow$ there are three possible $l$ values; 0, 1 and 2.
Given $m_{l}$ =1. $l$ can take the values 1 and 2.
$m_{s}$ can be $+ \frac{1}{2}$ and $- \frac{1}{2}$ .
Thus there can be only 4 electrons having the given quantum numbers.

Suggest Corrections

Similar questions

n = 3

m_{s} = - 1 / 2

n = 3

m_{s} = - \frac{1}{2}

n = 4, | m_{l} | = 1

m_{s} = - \frac{1}{2}

In an atom, the total number of electrons having quantum number
$n = 4, | m_{l} | = 1 a n d m_{s} = - \frac{1}{2} i s$ ___

n = 3

m_{s} = - \frac{1}{2}

Join BYJU'S Learning Program