Ch. 6 & 7 Review

1. Ch. 6 & 7 Review Light, Telescopes, & Spectroscopy

2. Question 1 • Which sequence of EMR is in order from the most energy to the least energy? • Microwaves, Visible Light, UV • UV, Red Light, Blue Light, Radio • X-Rays, UV, Red Light

3. Answer 1 • Choice #3… X-rays, UV, Red Light

4. Question 2 • Calculate the wavelength of a radio wave with a frequency of 90.1 Hz.

5. Answer 2 • 3.3 x 106 meters

6. Question 3 • How fast does an X-ray with a wavelength of 10 nm travel through empty space?

7. Answer 3 • Speed of Light… 3 x 108 m/s

8. Question 4 • Why do telescopes that collect visible light generally have a better resolving power than radio telescopes?

9. Answer 4 • Radio waves are low energy (have a long wavelength) so a telescope must collect more of them to get a better image. • That’s why most radio telescopes are interferometers.

10. Question 5 • Identify the type of telescope shown in the picture.

11. Answer 5 • Newtonian Focus

12. Question 6 • Identify the type of telescope in the picture.

13. Answer 6 • Cassegrain

14. Question 7 • Two stars are 0.045 arc seconds apart. You have a telescope with a 5.0 meter mirror. Are you able to see the two stars separately?

15. Answer 7 • Yes. The telescope has a resolving power of 0.023 arc seconds so it can see the two stars separately.

16. Question 8 • Calculate the magnifying power of a telescope with an objective focal length of 1.0 meter and an eyepiece focal length of 0.25 centimeters.

17. Answer 8 • 400 times

18. Question 9 • How much more light can a telescope with 10 meter diameter mirror collect than one with a 2 meter diameter mirror?

19. Answer 9 • 25 times

20. Question 10 • What type of spectrum is shown? • Explain why we see the bright lines?

21. Answer 10 • Emission Spectrum • The bright lines occur when the electrons in an excited gas drop down to their ground state energy level. As they return to the ground state, they emit photons of light at specific wavelengths.

22. Question 11 • One star has a surface temperature of 10,000 K and another has a surface temperature of 40,000 K. • How many times more energy is the hotter star emitting?

23. Answer 11 • 256 times • The 40,000 K star is 4 times as hot as the 10,000 K star. • T4… 44 = 256

24. Question 12 • At what wavelength would a star radiate the greatest amount of energy if it has a surface temperature of 6000 K? • Is that within the visible portion of the spectrum?

25. Answer 12 • 500 nm • Yes, the visible portion of the spectrum is from 400 – 700 nm.

26. Question 13 • Which jump requires the most energy? • Which jump would produce a photon with the longest wavelength?

27. Answer 13 • Most Energy -- #3 • Longest Wavelength -- #2

28. Question 14 • Identify the spectrum seen from each view point.

29. Answer 14 • Continuous – rainbow • Emission – bright lines • Absorption – dark lines

30. Question 15 • Which star is brighter, the top or the bottom? • How do you know?

31. Answer 15 • The spectral lines on the top spectrum are thinner. • Thinner lines mean lower density and brighter. • This is because brighter stars tend to be larger (in volume). The larger volume means a lower density (in general).

32. Question 16 • Which star is the coolest? • How do you know?

33. Answer 16 • I, the peak wavelength is longer than the other stars. • Longer wavelength means less energy means cooler star

34. Question 17 • The top spectrum is at rest. • Which spectrum shows a star that is traveling away from us at the fastest rate?

35. Answer 17 • Both spectra show red shift. • The bottom one shows more so it is traveling faster.

36. Question 18 • Antares and Betelgeuse are both M class stars. • Sirius and Vega are both A class stars. • Compare the temperature of these stars to the Sun.

37. Answer 18 • The Sun is a G2 star. • The class A stars are hotter than the Sun. • The class M stars are cooler than the Sun. • O,B,A,F,G,K,M

38. Question 19 • Calculate the Energy and the Frequency of one of the three most intense wavelengths absorbed in this spectrum. (3900 = 390nm)

39. Answer 19 • E = hc/λ and f = v/λ • 397.0 nm → 397 x 10-9 m E = 5.00 E-19 J and f =7.56 E14 Hz • 410.0 nm E = 4.85 E-19 J and f = 7.32 E14 Hz • 434.0 nm E = 4.58 E-19 J and f = 6.91 E14 Hz