Question

What is the maximum possible value of the angular momentum for an outer electron in the ground state of a bromine atom?

Short answer

Answer: The maximum possible value of the angular momentum for an outer electron in the ground state of a bromine atom is approximately 1.491964608 × 10⁻³⁴ Js.

Step-by-step solution

Find the ground state electron configuration of Bromine

Bromine has an atomic number of 35, meaning it has 35 electrons. The ground state electron configuration can be determined using the periodic table and the aufbau principle. For bromine, it is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

Identify the outer electron's quantum numbers

The outer electron is the one with the highest value of the principal quantum number (n). In the case of bromine, the outer electron is in the 4p orbital, which also has the highest energy level. The 4p orbital has the following quantum numbers: n=4, l=1, and ml can vary from -1 to 1.

Calculate the maximum possible angular momentum

The maximum possible value of the angular momentum (L) can be calculated using the formula: L = \sqrt{l (l +1)} \hbar Where l is the azimuthal quantum number, and \hbar is the reduced Planck constant (approximately 1.054571817 × 10⁻³⁴ Js). For the outer electron of a bromine atom, the maximum value of l is 1 (given by the 4p orbital).

Plug the value of l into the formula and find the maximum angular momentum

Now we can plug the maximum value of l (1) into the formula for angular momentum: L = \sqrt{1(1 + 1)} \hbar L = \sqrt{2} \hbar Thus, the maximum possible value of the angular momentum of the outer electron in the ground state of a bromine atom is sqrt(2) times the reduced Planck constant, which is approximately 1.491964608 × 10⁻³⁴ Js.