CHM 1045
Dr. Michael Blaber

Name_________________________

SS #__________________________

Homework # 6

  1. A green neon traffic light has a wavelength of 502 nm. What is the frequency of this light? (1 point)

    2. n * l = c

    n = c / l = (3.0 x 108 m/s) / (502 * 10-9 m) = 5.98 x 1014 s-1

     

  2. Given the following electromagnetic wave, draw beneath it a wave that has twice the frequency and half the amplitude (1 point)

  3. Your neighbor down the street decides to have laser hemorrhoid surgery. The medical laser has a wavelength of 2.3 x 10-7 meters. In a single shot it deposits 5.7 J of tissue-sizzling energy to the affected area. How many quanta of energy are this unfortunate individual absorbing? (2 points)

    4. n = c / l = (3.0 x 108 m/s) / (2.3 * 10-7 m) = 1.30 x 1015 s-1

    E = h * n = (6.63 x 10-34 J s) * (1.30 x 1015 s-1) = 8.62 x 10-19 J per quanta (1 point)

    5.7 J * (1 quanta / 8.62 x 10-19 J) = 6.61 x 1018 quanta (1 point)

  4. An electron transitions from the n = 3 to the n = 4 quantum state of the hydrogen atom. What is the wavelength of the associated photon? (1 points) Is the photon absorbed or emitted for the associated electron transition? (1 point). In what area of the electromagnetic spectrum is this located and is it visible light? (1 point)

    5. n = (RH/h)*(1/ni2 – 1/nf2) = (2.18 x 10-18 J / 6.63 x 10-34 J s)*(1/9 – 1/16) = 1.60 x 1014 s-1

    l = c / n = (3.0 x 108 m/s) / (1.60 x 1014 s-1) = 1.88 x 10-6 m or 1880 nm (1 point)

    n = +1.60 x 1014 s-1 positive value indicates light is absorbed (to excite the electron to a higher energy level) (1 point)

    l = 1880 nm , so this is in the infrared region of the spectrum (i.e. invisible) (1 point)

  5. Consider the wave behavior of matter: Dr. Blaber is loaded into a photon torpedo and ejected from the Starship Enterprise at 9/10th’s the speed of light for making fun of the captain’s hemorrhoids. What is Dr. Blaber’s characteristic wavelength? (Dr. Blaber is a svelte 195 pounds) (1 point)

    6. l = h/(mv) = (6.63 x 10-34 J s)/{(195 lbs*0.454 kg/lb)*(0.9*3 x 108 m/s)} =

    = 2.77 x 10-44 J kg-1 m-1 s2 * (1 kg m2 s-2 / J) = 2.77 x 10-44 m

  6. What is the designation for the electron subshell with principle quantum number = 4 and azimuthal quantum number = 2 ? (1 point)

    7. n = 4, l = 2

    4d

  7. How many orbitals are there in the subshell with a principle quantum number of 3, and an azimuthal quantum number of 2 (1 point)

n = 3

l = 2

3d, which has 5