Find the maximum possible magnitude for the orbital angular momentum of an electron in the \(n=7\) state of hydrogen.

Short Answer

Expert verified
The maximum possible magnitude of the orbital angular momentum for an electron in the \(n=7\) state of hydrogen = \(\sqrt{42}\hbar\).

Step by step solution

01

Identify the value of the principal quantum number

The principal quantum number 'n' is given in the problem statement. In this case, 'n' is equal to 7.
02

Calculate the azimuthal quantum number 'l'

The azimuthal quantum number 'l' is equal to n-1. Here, since 'n' is 7, 'l' equals 7 - 1 = 6.
03

Use the angular momentum formula

Substitute the azimuthal quantum number 6 into the formula for the magnitude of the angular momentum \[L=\sqrt{l(l+1)}\hbar \]. Computing, we get \[L=\sqrt{6(6+1)}\hbar =\sqrt{42}\hbar\].
04

Simplify the expression

Simplifying the result, we are left with the final answer for the maximum possible magnitude of the orbital angular momentum.

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