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Chp 5 problems relay

Problem 1 If the wavelength is 5.00 x 10¯ 7 m, determine the frequency of this light. Also n ame the area on the electromagnetic spectra that contains this frequency light? (you may use your book.). Chp 5 problems relay.

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Chp 5 problems relay

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  1. Problem 1 If the wavelength is 5.00 x 10¯7 m, determine the frequency of this light.Also name the area on the electromagnetic spectra that contains this frequency light?(you may use your book.) Chp 5 problems relay

  2. Problem 2 Do the following conversions a) 500.0 nm = _____m b) 98.3 MHz = _________Hz Chp 5 problems relay

  3. Problem 3If it takes 3.36 x 10-19 J of energy to eject an electron from the surface of a certain metal, calculate the wavelength of light that can ionize the metal. Is this phenomena explained by the wave or particle nature of light? What is the name of this phenomena? Chp 5 problems relay

  4. Read the section and/or the table below to help answer problem 4Spectral TypesStars are divided into groups called spectral types (also called spectral classes) which are based on the strength of the hydrogen absorption lines. The A-type stars have the strongest hydrogen lines, B-type next strongest, F-type next, etc. Astronomers discovered that the line strengths also depend on the temperature. The presence of other atomic or ion lines are used in conjunction with the hydrogen spectrum to determine the particular temperature of the star. After some rearranging and merging of some classes, the spectral type sequence is now OBAFGKM when ordered by temperature. The O-type stars are the hottest stars and the M-type stars are the coolest. Each spectral type is subdivided into 10 intervals, e.g., G2 or F5, with 0 hotter than 1, 1 hotter than 2, etc. About 90% of the stars are called main sequence stars. The other 10% are either red giants, supergiants, white dwarfs, proto-stars, neutron stars, or black holes. The table below gives some basic characteristics of the different spectral classes of main sequence stars. Notice the trends in the table: as the temperature of the main sequence star increases, the mass and size increase. Also, because of the relation between luminosity and the size and temperature of a star, hotter main sequence stars are more luminous than cooler main sequence stars. However, there are limits to how hot a star will be, or how massive and large it can be. Understanding why the constraints exist is the key to understanding how stars work. Chp 5 problems relay 4) Two stars have equal strengths of their hydrogen lines. Star A has lines from helium present while star B has lines of ionized calcium present. Which colors might they appear to be and which star is hotter? Explain your reasoning.

  5. 5) Explain what is happening and the data being gathered. What do you know about the frequency and energy of the photons of the different colors? • Chp 5 problems relay

  6. Chp 5 problems relay 6. Draw the following • the electron configuration for calcium • orbital diagram for sulfur • noble-gas notation for germanium • Lewis (electron)-dot structure for nitrogen

  7. Prob 1) SolutionIf the wavelength is 5.00 x 10¯7 m, then determine the frequency of the light.3 x 108/ 5x 10-7v= 6.00 x 1014s¯1 or Hz or 1/sInfrared Chp 5 problems relay

  8. Prob 2 solution nano= 10 -9 and Mega is 106a) 500.0 x 10¯9 m or5.000 x 10¯7 mb)98.3 x 106 Hz or 9.83 x 107Hz Chp 5 problems relay

  9. Problem 3) SolutionDetermine the frequency: E = hν3.36 x 10-19 J = (6.626 x 10¯34 J s) (x) x = 5.071 x 1014 s¯1Determine the wavelength: λν= c (x) (5.071 x 1014 s¯1) = 3.00 x 108 m/s 3.00 x 108 m/sx = 5.071 x 1014s¯1x = 5.916 x 10¯7 m or 592 nm particle theory , photoelectric effect. Chp 5 problems relay

  10. 4) Solution A will be bluest and B will be white to yellow to red depending on its exact temperature.A is hotter as the table shows helium stars at 40,000 and calcium stars at 7,000 or lower. Chp 5 problems relay

  11. 5) Solution Answers should include that E=hv and c=lv. Also they should explain that a single electron is jumping to various shells (orbitals) and falling back down to ground state and emitting the spectra shown. (hydrogen)

  12. 6. Solution • the electron configuration for calcium (20) 1s22s22p63s23p64s2 b. orbital diagram for sulfur (16) ___ ___ ___ ___ ___ ______ ___ ___ 1s 2s 2p 3s 3p c. noble-gas notation for germanium (32) [Ar]4s23d104p2 d. Lewis (electron)-dot structure for nitrogen All of these are acceptable

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