Exam 1-solutions

# Exam 1-solutions - Version 292 – Exam 1 – mccord...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 292 – Exam 1 – mccord – (51600) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 200 nm photon has how many times the energy of a 700 nm photon? 1. 0.24 2. 4.2 3. 3.5 correct 4. 9 . 93 × 10 − 19 5. 2 . 84 × 10 − 19 6. 0.29 Explanation: Energy of Light: E = hc λ For the 200 nm photon: E = hc λ = (6 . 626 × 10 − 34 J · s)(3 × 10 8 m · s − 1 ) 200 × 10 − 9 m = 9 . 94 × 10 − 19 J For the 700 nm photon: E = hc λ = (6 . 626 × 10 − 34 J · s)(3 × 10 8 m · s − 1 ) 700 × 10 − 9 m = 2 . 84 × 10 − 19 J Thus 9 . 94 × 10 − 19 J 2 . 84 × 10 − 19 J = 3 . 5 002 10.0 points Carbon emits photons at 745 nm when ex- posed to blackbody radiation. How much energy would be obtained if 44 g of carbon were irradiated? 1. 2 . 67 × 10 − 19 J 2. 5 . 90 × 10 5 J correct 3. 7 . 08 × 10 3 J 4. 9 . 11 × 10 − 21 J 5. 7 . 08 × 10 6 J 6. 1 . 17 × 10 − 17 J Explanation: λ = 745 nm = 7 . 45 × 10 − 7 m m C = 44 g Assume each carbon atom emits one pho- ton. For each photon E 1 = h ν = h c λ = (6 . 626 × 10 − 34 J · s) (3 × 10 8 m / s) 7 . 45 × 10 − 7 m = 2 . 66819 × 10 − 19 J , so the total energy emitted is E = E 1 n = 2 . 66819 × 10 − 19 J C atom × 6 . 022 × 10 23 C atoms 1 mol C atoms × 1 mol C atoms 12 . 01 g C × (44 g C) = 5 . 88663 × 10 5 J . 003 10.0 points Which of the following emission lines corre- sponds to part of the Balmer series of lines in the spectrum of a hydrogen atom? A) n = 2 → n = 1 B) n = 4 → n = 2 C) n = 4 → n = 1 D) n = 3 → n = 2 E) n = 4 → n = 3 1. B and D only correct 2. E only 3. A, D, and E only Version 292 – Exam 1 – mccord – (51600) 2 4. B, C, and E only 5. D and E only 6. B and C only 7. A and C only Explanation: The Balmer series is produced by elec- tronic transitions which either begin (absorp- tion spectra) or end (emission spectra) at the energy level n = 2. These correspond mostly to the visible region. 004 10.0 points Bohr’s theory of the hydrogen atom assumed that 1. because a hydrogen atom has only one electron, the emission spectrum of hydrogen should consist of only one line. 2. the electron can exist in any one of a set of discrete states (energy levels) and can move from one to another by emitting or absorbing radiation. correct 3. the electron in a hydrogen atom can jump from one energy level to another without gain or loss of energy. 4. energy, in the form of radiation, must be continually supplied to keep the electron moving in its orbit. 5. electromagnetic radiation is given off when the electrons move in an orbit around the nucleus....
View Full Document

{[ snackBarMessage ]}

### Page1 / 10

Exam 1-solutions - Version 292 – Exam 1 – mccord...

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

View Full Document
Ask a homework question - tutors are online