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Chapter 36: Problem 20

A hydrogen atom is in the \(6 f\) state. Find (a) its energy and (b) the magnitude of its orbital angular momentum.

Short Answer

Expert verified
The energy of the hydrogen atom is \(-0.378 eV\) and magnitude of its orbital angular momentum is \(2 \sqrt {3} \hbar\).

Step by step solution

01

Identify Given Values

In this problem, it is known that the hydrogen atom is in the \(6f\) state. This means the principal quantum number \(n\) is 6 and the azimuthal quantum number \(l\) is 3 (as letter \(f\) corresponds to \(l=3\)).
02

Calculate Energy

The energy of a hydrogen atom in quantum physics can be calculated by the energy equation: \[E = - \frac {13.6 eV} {n^2} \] Substituting the given value of \(n\), the energy becomes \[E = - \frac {13.6 eV} {6^2} = -0.378 eV \]
03

Calculate Orbital Angular Momentum

For a hydrogen atom, the magnitude of the orbital angular momentum \(L\) can be calculated using the formula \[ L = \sqrt {l(l + 1)} \hbar \] Substituting the given value of \(l\) gives \[ L = \sqrt {3(3 + 1)} \hbar = 2 \sqrt {3} \hbar \]

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