For an electron in the ground state of a hydrogen atom, which set of quantum numbers is correct?

medium MCQ 2 marks Class 12 · Physics · Atoms

A. $n = 1, l = 1, m_{l} = 0, m_{s} = + \frac{1}{2}$
B. $n = 1, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}$
C. $n = 1, l = 0, m_{l} = 1, m_{s} = - \frac{1}{2}$
D. $n = 2, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}$

Answer

n = 1, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}

The correct answer is B. $n = 1, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}$ — Hence the only consistent set is $n = 1, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}$ .

Step-by-step solution

The principal quantum number for the ground state is $n = 1$ .
For a given $n$ , the orbital quantum number $l$ can take values $0, 1, \dots, n - 1$ ; therefore $l = 0$ for $n = 1$ .
The magnetic quantum number $m_{l}$ ranges from $- l$ to $+ l$ in integer steps; with $l = 0$ , $m_{l} = 0$ .
The spin quantum number $m_{s}$ can be $+ \frac{1}{2}$ or $- \frac{1}{2}$ ; either value is allowed, and the conventional choice is $+ \frac{1}{2}$ .
Hence the only consistent set is $n = 1, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}$ .

Common mistake: Assuming

l = 1

for

n = 1

, which violates the rule

l < n

Concept tested: quantum numbers of hydrogen ground state

Published 15 September 2026 · How this solution was prepared and checked

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