Question

instructions: no books, papers, or notes of any kind permitted. a scientific calculator is permitted on this quiz. an si prefix table, a greek prefix table, and a table of electromagnetic radiation are provided. a periodic table is supplied. you will have up to 15 minutes to complete this quiz. multiple choice questions: task: clearly select or circle the letter for the answer that best fits the question. criteria: each correct answer is worth 2 points. q1. how many valence electrons are in the sulfur atom? a) 4. b) 6. c) 8. d) 16. q2. what is the frequency of orange light with a wavelength of 590 nanometers? a) 5.90×10¹⁴ hz. b) 5.90×10⁷ hz. c) 5.08×10¹⁴ hz. d) 5.08×10⁻¹⁹ hz. q3. a green photon is able to excite an electron from n=1 to n=3 in a certain atom. an excitation from n=3 to n=2 would involve: a) absorption of a blue photon. b) emission of a green photon. c) emission of a red photon. d) absorption of a red photon. p.t.o. for remaining quiz questions...

Explanation:

Step1: Recall sulfur's electron - configuration

Sulfur has an atomic number of 16. Its electron - configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}\). The outermost shell is the third shell (\(n = 3\)), and the number of electrons in the \(3s\) and \(3p\) sub - shells is \(2 + 4=6\). So, the number of valence electrons in a sulfur atom is 6.

Step2: Use the speed - of - light formula for frequency calculation

The speed of light \(c = \lambda
u\), where \(c = 3\times10^{8}\ m/s\), \(\lambda\) is the wavelength, and \(
u\) is the frequency. Given \(\lambda=590\ nm = 590\times10^{-9}\ m\). Then \(
u=\frac{c}{\lambda}=\frac{3\times10^{8}}{590\times10^{-9}}\approx5.08\times 10^{14}\ Hz\).

Step3: Analyze electron transition and photon energy

When an electron goes from a higher energy level (\(n = 3\)) to a lower energy level (\(n = 2\)), it emits a photon. The energy of the photon is given by \(E = h
u\). The energy difference between \(n = 3\) and \(n = 2\) is less than the energy difference between \(n = 1\) and \(n = 3\). Since energy is proportional to frequency and blue photons have higher frequencies (more energy) than green photons and red photons have lower frequencies (less energy) than green photons, the transition from \(n = 3\) to \(n = 2\) involves the emission of a red photon.

Answer:

Q1. B. 6
Q2. C. \(5.08\times10^{14}\ Hz\)
Q3. C. Emission of a red photon.