The electrons, identified by quantum number $n\ and\ l$, $n = 3;l = 2$ $n = 5;l = 0$ $n = 4;l = 1$ $n = 4;l = 2$ $n = 4;l = 0$ can be placed in order of increasing energy, as

Question

Dr. Abubacker · Accepted Answer

(c)
Higher the value of $(n + l)$, higher will be the energy of electrons. If value of $(n + l)$ is same for any two or more electrons, the electron with higher value of $n$, has higher energy. Hence, the correct order of energy is
$V < I < III < II < IV$
$\because\ \ \ (n + 1)$ 4 5 5 5 6

Q. The electrons, identified by quantum number $n\ and\ l$, $n = 3;l = 2$ $n = 5;l = 0$ $n = 4;l = 1$ $n = 4;l = 2$ $n = 4;l = 0$ can be placed in order of increasing energy, as

Explanation

Reviewed by Dr. Abubacker

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