Which one of the following is the correct set | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rules for quantum numbers are as follows:$1$. Principal quantum number $n$ can be any positive integer $(1, 2, 3, ...)$.$2$. Azimuthal quantum

Vedclass · Accepted Answer

$(4, 3, -2, -\frac{1}{2})$

Vedclass · Answer

(B) The rules for quantum numbers are as follows:$1$. Principal quantum number $n$ can be any positive integer $(1, 2, 3, ...)$.$2$. Azimuthal quantum number $\ell$ ranges from $0$ to $n-1$.$3$. Magnetic quantum number $m$ ranges from $-\ell$ to $+\ell$.$4$. Spin quantum number $s$ can be $+\frac{1}{2}$ or $-\frac{1}{2}$.Evaluating the options:$A$: $(3, 0, -1, +\frac{1}{2})$ is incorrect because $m$ cannot be $-1$ when $\ell=0$.$B$: $(4, 3, -2, -\frac{1}{2})$ is correct because for $n=4$, $\ell=3$ is allowed, and for $\ell=3$, $m=-2$ is allowed.$C$: $(3, 1, -2, -\frac{1}{2})$ is incorrect because $m$ cannot be $-2$ when $\ell=1$.$D$: $(4, 2, -3, +\frac{1}{2})$ is incorrect because $m$ cannot be $-3$ when $\ell=2$.

Which one of the following is the correct set of four quantum numbers $(n, l, m, s)$?

Similar Questions

Statement $A$: In a potassium atom,the $19^{th}$ electron enters the $4s$-orbital instead of the $3d$-orbital.
Reason $R$: The $(n + l)$ rule is followed to determine the minimum energy of an orbital.

What is an atomic orbital?

The graph between $|\psi|^2$ and $r$ (radial distance) is shown below. This represents

In an atom,the total number of electrons having quantum numbers $n=4$,$|m_{\ell}|=1$ and $m_s=-1/2$ is

The correct decreasing order of energy for the orbitals having the following set of quantum numbers:
$A$. $n=3, \ell=0, m=0$
$B$. $n=4, \ell=0, m=0$
$C$. $n=3, \ell=1, m=0$
$D$. $n=3, \ell=2, m=1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module