Homework 11 - montes(dam2772 – Homework11 – shih –(56310 1 This print-out should have 20 questions Multiple-choice questions may continue on

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: montes (dam2772) – Homework11 – shih – (56310) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For the following types radio waves infared radiation visible light ultraviolet radiation gamma radiation of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes from the top of the list to the bottom? Wavelength Frequency Photon Energy 1. Decreases Decreases Increases 2. Increases Decreases Increases 3. Increases Decreases Decreases 4. Decreases Increases Increases cor- rect 5. Increases Increases Increases Explanation: The wavelength decreases from the top of the list to the bottom, so the frequency f = v λ increases, as does and the photon energy E = h f . 002 (part 1 of 3) 10.0 points Consider the four lowest energy levels shown in the diagram of a certain atom. n = 1 n = 2 n = 3 n = 4 How many spectral lines will result from all possible transitions among these levels? 1. 8 2. None of these 3. 4 4. 5 5. 6 correct Explanation: n = 1 n = 2 n = 3 n = 4 003 (part 2 of 3) 10.0 points Which transistion corresponds to the lowest- frequency light emitted? 1. n = 4 to n = 1 2. None of these 3. n = 3 to n = 2 4. n = 4 to n = 3 correct 5. n = 4 to n = 2 Explanation: The lowest frequency occurs going from n = 4 to n = 3. 004 (part 3 of 3) 10.0 points Which transistion corresponds to the highest- frequency light emitted? 1. n = 3 to n = 2 2. n = 4 to n = 3 3. None of these 4. n = 4 to n = 2 montes (dam2772) – Homework11 – shih – (56310) 2 5. n = 4 to n = 1 correct Explanation: The highest frequency occurs going from n = 4 to n = 1. keywords: 005 (part 1 of 3) 10.0 points An electron is in the second Bohr orbit of hydrogen. Find the speed of the electron. The value of ¯ h is 1 . 05 × 10- 34 J · s . The Bohr radius is 5 . 29 × 10- 11 m. Correct answer: 1 . 08939 × 10 6 m / s. Explanation: Let : n = 2 , ¯ h = 1 . 05 × 10- 34 J · s , m = 9 . 11 × 10- 31 kg , and a = 5 . 29 × 10- 11 m (Bohr radius) ....
View Full Document

This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.

Page1 / 6

Homework 11 - montes(dam2772 – Homework11 – shih –(56310 1 This print-out should have 20 questions Multiple-choice questions may continue on

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