Chapter 8: Problem 62

The valence electron of potassium is excited to a \(5 d\) state. (a) What is the magnitude of the electron's orbital angular momentum? (b) How many states are possible along a chosen direction?

Short Answer

Expert verified

The magnitude of the electron's orbital angular momentum is \(\sqrt{6}\hbar\) and there are 5 possible states along a chosen direction.

Step by step solution

Calculate the magnitude of the orbital angular momentum

The magnitude of the electron's orbital angular momentum can be calculated using the formula \(L = \sqrt{l(l+1)}\hbar\) where \(l\) is the orbital quantum number (it is equal to 2 for a \(d\) orbital) and \(\hbar\) is the reduced Plank's constant. Substituting the value of \(l\) into the formula, we have \(L = \sqrt{2(2+1)}\hbar = \sqrt{6}\hbar\)

Calculate the number of possible states

The number of possible states along a chosen direction (for example, the z-direction) is given by the different possible values of the magnetic quantum number \(m\). For a given \(l\) value, \(m\) can take on values ranging from \(-l\) to \(l\), inclusive. Hence \(m\) can be any integer from -2 to 2, since \(l = 2\). Thus, there are \(2 * 2 + 1 = 5\) different possible states.

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The valence electron of potassium is excited to a \(5 d\) state. (a) What is the magnitude of the electron's orbital angular momentum? (b) How many states are possible along a chosen direction?

Short Answer

Step by step solution

Calculate the magnitude of the orbital angular momentum

Calculate the number of possible states

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