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Chapter 37: Problem 29

A molecule drops from the \(l=2\) to the \(l=1\) rotational level, emitting a 2.50 -meV photon. If the molecule then drops to the rotational ground state, what energy photon will it emit?

Short Answer

Expert verified
The photon emitted when the molecule drops to the ground state will also have an energy of \(2.50 \times 10^{-3}\) meV.

Step by step solution

01

Understanding Rotational Energy Levels

In quantum physics, the rotational energy levels of a diatomic molecule are given by \(E = l(l + 1) \times \hbar^2 / 2 \mu r^2\), where \(l\) is the rotational quantum number, \(\hbar\) is the reduced Planck's constant, \(\mu\) is the reduced mass of the molecule and \(r\) is the bond length between the atoms in the molecule. When a molecule moves from a higher energy level to a lower energy level, it emits a photon with energy equal to the difference between the energy levels.
02

Finding the Change in Energy for the First Transition

The given problem states that the transition is from \(l = 2\) to \(l = 1\), so the change in energy \(\Delta E\) will be the difference between the energies of these two levels. Using the formula for rotational energy levels, we have \(\Delta E = E_1 - E_2 = (2(2+1) - 1(1+1)) \times \hbar^2 / 2 \mu r^2 = \hbar^2 / \mu r^2\). This energy difference corresponds to the energy of the emitted photon, which is given as 2.50 meV.
03

Correlating the Energy Difference with the Emitted Photon Energy

Since the energy of the emitted photon is equal to the energy difference between the rotational states, we have \(\hbar^2 / \mu r^2 = 2.50 \times 10^{-3}\) eV. Note that 1 eV = 1.6 × 10^{-19} J, so the energy difference in Joules is \(2.50 \times 10^{-3} \times 1.6 × 10^{-19} J = 4 \times 10^{-22} J\). We can now put this value back into the equation we used to get the energy difference, and solve for \(\hbar^2 / \mu r^2\).
04

Finding the Energy of the Second Photon

Now the molecule drops from \(l = 1\) to the ground state (\(l = 0\)). Using the rotational energy formula, the energy difference is now \(E_1 - E_0 = 1(1+1) - 0(0+1) \times \hbar^2 / 2 \mu r^2 = \hbar^2 / \mu r^2\), which is equal to the value determined in Step 3. So, the energy of the second emitted photon is also \(4 \times 10^{-22} J\). Converting to eV yields \(4 \times 10^{-22} J / 1.6 × 10^{-19} J/eV = 2.50 \times 10^{-3}\) meV.

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