Question

Give the maximum number of electrons in an atom that can have these quantum numbers: a. $n=0,\ell =0,m_{\ell }=0$ b. $n=2,\ell =1,m_{\ell }=-1,m_s=-\frac{1}{2}$ c. $n=3,m_s=+\frac{1}{2}$ d. $n=2,\ell =2$ e. $n=1,\ell =0,m_{\ell }=0$

Short answer

a. 0 electrons b. 1 electron c. 9 electrons d. 0 electrons e. 2 electrons

Step-by-step solution

a. $n=0$, $\ell =0$, $m_{\ell }=0$

The principal quantum number n must be a positive integer (n > 0), but here n is equal to 0. Therefore, this quantum number set is not valid, and there are no electrons associated with these quantum numbers.

b. $n=2$, $\ell =1$, $m_{\ell }=-1$, $m_s=-\frac{1}{2}$

These quantum numbers are valid according to the rules. Since the spin quantum number (m_s) is given and uniquely defines an electron, there can be only one electron with these quantum numbers in an atom.

c. $n=3$, $m_s=+\frac{1}{2}$

In this case, the angular momentum quantum number ($\ell$) and magnetic quantum number ($m_{\ell }$) are not provided. Based on the rules, for a given principal quantum number n, there can be $n^2$ orbitals and 2 electrons per orbital (one for each spin). For n=3, there are $3^2=9$ orbitals. Since $m_s$ is specified, each orbital can accommodate only one electron with this spin. Therefore, the maximum number of electrons with these quantum numbers is 9.

d. $n=2$, $\ell =2$

The angular momentum quantum number ($\ell$) has a maximum value of n-1. In this case, $\ell$ is equal to n, which is not valid. Thus, there are no electrons associated with these quantum numbers.

e. $n=1$, $\ell =0$, $m_{\ell }=0$

These quantum numbers are valid according to the rules. Since the magnetic quantum number (m_ell) and the angular momentum quantum number (l) are both given and zero, there is only one orbital associated with these quantum numbers. Each orbital can accommodate 2 electrons (one with spin +1/2 and one with spin -1/2). Therefore, the maximum number of electrons with these quantum numbers is 2.