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Chapter 36: Problem 39

A hydrogen atom is in an \(l=2\) state. What are the possible angles its orbital angular momentum vector can make with a given axis?

Short Answer

Expert verified
The possible angles that the orbital angular momentum vector can make with a given axis for a hydrogen atom in an \(l=2\) state are found to be: \(\cos^{-1}\) (-1), \(\cos^{-1}\) (-0.5), \(\cos^{-1}\)(0), \(\cos^{-1}\)(0.5), and \(\cos^{-1}\)(1), which can be calculated to obtain the final values in degrees or radians, as needed.

Step by step solution

01

Understanding the Possible Values of \(m_l\)

Start by identifying that possible values for \(\(m_l\)\) can be any integer from \(-l\) to \(+l\), inclusive. In this case, \(\(l=2\)\), so \(\(m_l\)\) can be \(-2, -1, 0, 1, 2\).
02

Calculating the Corresponsing Angle

Next, calculate the corresponding angle \(\theta\) that each of these values for \(\(m_l\)\) would produce, using the relationship \(\(m_l = l \cos(\theta)\)\). Re-arranging this gives \(\cos(\theta)\) = \(\(m_l/l\)\), which can then be converted to an angle via the inverse cosine function.
03

Calculating the Angles for Each Value of \(m_l\)

Now, calculate the angle corresponding to each value of \(\(m_l\)\). Doing this yields the possible angles as: \(\cos^{-1}\) (-1), \(\cos^{-1}\) (-0.5), \(\cos^{-1}\)(0), \(\cos^{-1}\)(0.5), and \(\cos^{-1}\)(1). Calculate these values to get the desired angle in degrees or radians, as per preference. This will give the possible angles the orbital angular momentum vector can make with a given axis.

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Most popular questions from this chapter

Which of the following is not a possible value for the magnitude of the orbital angular momentum in hydrogen: (a) \(\sqrt{12} \hbar\) (b) \(\sqrt{20} \hbar ;\) (c) \(\sqrt{30} \hbar ;\) (d) \(\sqrt{40} \hbar ;\) (e) \(\sqrt{56} \hbar ?\)

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