Answer:
[tex]L_{max} = 1.414 [/tex] ℏ
Given:
Principle quantum number, n = 2
Solution:
To calculate the maximum angular momentum, [tex]L_{max}[/tex], we have:
[tex]L_{max} = \sqrt {l(1 + l)}[/tex] (1)
where,
l = azimuthal quantum number or angular momentum quantum number
Also,
n = 1 + l
2 = 1 + l
l = 1
Now,
Using the value of l = 1 in eqn (1), we get:
[tex]L_{max} = \sqrt {1(1 + 1)} = \sqrt 2[/tex]
[tex]L_{max} = 1.414 [/tex] ℏ
