For $n = 3$ energy level,the number of possible orbitals (all kinds) are

How many electrons in an atom may have the following quantum numbers?
$(a)$ $n=4, m_{s}=-\frac{1}{2}$
$(b)$ $n=3, l=0$

Define the principal quantum number $(n)$.

For electrons having principal quantum number $n = 3$,the number of $(i)$ subshells and $(ii)$ orbitals would be respectively:

Match List-$I$ with List-$II$ :

List-$I$ (Quantum number)	List-$II$ (Orbital)
$a. n=2, \ell=1$	$i. 2s$
$b. n=3, \ell=2$	$ii. 3s$
$c. n=3, \ell=0$	$iii. 2p$
$d. n=2, \ell=0$	$iv. 3d$

Choose the correct answer from the options given below :

Which of the following statement$(s)$ is/are true?
$A$. If two orbitals have the same value of $(n+l)$, the orbital with lower value of $n$ will have lower energy.
$B$. Energies of the orbitals in the same subshell increase with increase in atomic number.
$C$. The size of $2p_x$ orbital is less than the size of $3p_x$ orbital.
$D$. Among $5f, 6s, 4d, 5p$ and $5d$ orbitals, none of the orbitals have $2$ radial nodes.