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The quantum numbers of four electrons (e1 to e4) are given below :
$\begin{array}{ccccr} n & l & m & s \\ e 1 & 3 & 0 & 0 & + \frac{1}{2} \\ e 2 & 4 & 0 & 0 & \frac{1}{2} \\ e 3 & 3 & 2 & 2 & - \frac{1}{2} \\ e 4 & 3 & 1 & - 1 & \frac{1}{2} \end{array}$
The correct order of decreasing energy of these electrons is:

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\begin{aligned} e 4 > e 3 > e 2 > e 1 \end{aligned}

e 2 > e 3 > e 4 > e 1

e 3 > e 2 > e 4 > e 1

none of these

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detailed solution

Correct option is C

the energy formula for electron in quantum number is $\propto (n + 1)$ value

$\begin{aligned} e 1 = 3 + 0 = 3 \\ e 2 = 4 + 0 = 4 \\ e 3 = 3 + 2 = 5 \\ e 4 = 3 + 1 = 4 \end{aligned}$

If $(n + 1)$ value is same, then larger n value has more energy.

so,

correct order of decreasing energy of the electrons is $e 3 > e 2 > e 4 > e 1$ .

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The quantum numbers of four electrons (e1 to e4) are given below : nlmse1300+12e240012e3322−12e431−112The correct order of decreasing energy of these electrons is: