view-notes-pcs110-013 - CHAPTER 42 MOLECULAR AND...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 42 MOLECULAR AND SOLID-STATE PHYSICS Section 42- 2: Molecular Energy Levels Problem 1. Find the energies of the first four rotational states of the HCl molecules described in Example 42-1. Solution The energies of rotational states (above the j = 0 state) are given by Equation 42-2, where for the HCl molecule, 2 2 63 = I = . meV (from Example 42-1). Thus, E j j I j j rot meV. = + = + ( ) ( ) . 1 2 1 2 63 2 1 2 = For j = 1 2 , , , and 3 2 63 7 89 15 78 , , . , . , . . meV meV and meV rot E = Problem 2. Find the wavelength of electromagnetic radiation needed to excite oxygen molecules (O 2 ) to their first rotational excited state. The rotational inertia of an oxygen molecule is 1 95 10 46 2 . . - kg m Solution The difference in energy between the j = 1 and j = 0 states is E I I = +- = 1 1 1 2 2 2 ( )( ) . = = The photon wavelength corresponding to this transition is l p p = = = = - hc E hcI cI = = = = 2 8 46 2 2 2 3 10 1 055 ( / ) ( . m s)(1.95 10 kg m 10 3 34- = J s mm. ) .48 Problem 3 A molecule drops from the j = 2 to the j = 1 rotational level, emitting a 2.50-meV photon. If the molecule then drops to the rotational ground state, what energy photon will be emitted? Solution The energy of a photon emitted in a transition between rotational levels with j = - 1 is shown in Example 42-1 to be E j I j j - = ( ) . 1 2 = Using the given data for the first transition, j j = = 2 1 to , we find 2 3 2 50 10 2 = = I = =- . eV 1 25 10 3 . - eV. This is also equal to the energy of a photon in the j j = = 1 to transition. Problem 4. Calculate the wavelength of a photon emitted in the j = 5 to j = 4 transition of a molecule whose rotational inertia is 1 75 10 47 2 . . - kg m Solution The energy difference between adjacent rotational levels is proportional to the upper j-value (see Example 42-1 and the solution to Problem 6 below), so l p p = = = =-- hc E cI j = = = 2 2 3 10 1 75 10 5 1 055 10 8 47 34 ( / . ) ( . ) m s)( kg m J s 2 62 5 . m. m Problem 5. Photons of wavelength 1.68 cm excite transitions from the rotational ground state to the first rotational excited state in a gas. What is the rotational inertia of the gas molecules? Solution The energy of the absorbed photon equals the difference in energy between the j = 1 and j = 0 rotational levels, which is (see Example 42-1) 2 1 5 23 1240 1 68 7 38 10 118 10 = = = I E hc = = = = = -- l eV nm cm eV J. . . . Therefore, I = = --- ( . ( . ) .41 . 1 055 10 118 10 9 10 34 2 23 46 J s) J kg m 2 = CHAPTER 42 979 Problem 6. A molecule absorbs a photon and jumps to the next higher rotational state. If the photon energy is three times what would be needed for a transition from the rotational ground state to the first rotational excited state, between what two levels is the transition?...
View Full Document

This note was uploaded on 10/19/2010 for the course PHY 78647 taught by Professor Miller during the Spring '09 term at Albany College of Pharmacy and Health Sciences.

Page1 / 13

view-notes-pcs110-013 - CHAPTER 42 MOLECULAR AND...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online