An atom has completely filled inner shells and a single valence electron in an excited p state. The filled inner shells have an orbital momentum equal to zero. What is the magnitude of the orbital angular momentum of the atom?
A) 1.0 $\frac { h } { 2 \pi }$
B) 1.2 $\frac { h } { 2 \pi }$
C) 1.4 $\frac { h } { 2 \pi }$
D) 1.7 $\frac { h } { 2 \pi }$
E) 2.0 $\frac { h } { 2 \pi }$

Question

Quizplus · Accepted Answer

The magnitude of the orbital angular momentum of an electron in an excited state can be calculated using the formula:
L = √(l(l+1) * ħ^2)
where l is the orbital quantum number and ħ is the reduced Planck constant. 
Since the valence electron is in an excited p state, l = 1. 
Thus, 
L = √(1(1+1) * ħ^2) 
= √2 * ħ
= 1.414 * ħ
= 1.4 * (h/2π)
Therefore, the correct answer is C.

An Atom Has Completely Filled Inner Shells and a Single h2π\frac { h } { 2 \pi }2πh​

An Atom Has Completely Filled Inner Shells and a Single $\frac { h } { 2 \pi }$