Lone Star CHEM 1411 EXAM 3 Fall 2017. Work 33 of 36 for 3 PTS ea. Additional 3 f

Question

Lone Star CHEM 1411 EXAM 3 Fall 2017. Work 33 of 36 for 3 PTS ea. Additional 3 for 2 bonus PTS each. 1. What is the wavelength of a photon having a frequency of 4.50 x10" Ha? ( 3.00 10 m/s, h-6.63x 10 J s, 1 m 1 x10 n m) A) 667 nm B) 1.50 x 10nm C) 4.42 x 10 nm D) 0.0895 nm E) 298 x10". nm 2. 79 kJ/mol? (e -3.00 x10 m/s, What is the wavelength of photons that have molar energy of 4 h 6.63 x10* J s, N 6.02 x10 mol. 1m-1 x10nm) A) 1.20 x 10* nm B) 7.96 x 101 m C) 250 nm D) 4.15 x 10 m E) 2.50 x 10*m 3. In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus, A) energy is absorbed. B) light is emitted. C) energy is emitted. D) no change in energy occurs. E) none of these What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition fron level n-6 to level n-17 (c-3.00 x 10·m/s, h = 6.63 × 1044 J-s, A, = 2.179 10""D 4. A) 9.39 x 10-*m B) 7.06 × 10-27m C) 1.07 × 107 m D) 2.12 x 10-1"m E) 3.20x 105m

Explanation / Answer

Q1

relate WL via:

WL = c/v

WL = (3*10^8)/(4.5*10^14)

WL = 6.66*10^-7 --> 666 nm

choose A

Q2

E = 479 kJ/mol --> 479000 J/mol --> 479000 * 1/(6.022*10^23) = 7.954*10^-19

WL = hc/E

WL = (6.63*10^-34)(3*10^8)/(7.954*10^-19)

WL = 2.5006*10^-7 m

WL = 250 nm

Q3

if more distant, this is "higher" in energy therefore, it must absorb energy

Q4

Apply Rydberg Formula

E = R*(1/nf^2 – 1/ni ^2)

R = -2.178*10^-18 J

Nf = final stage/level

Ni = initial stage/level

E = Energy per unit (i.e. J/photon)

E = (-2.178*10^-18)*(1/1^2 – 1/6 ^2)

E = 2.11*10^-18

For the wavelength:

WL = h c / E

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = wavelength in meters

WL = (6.626*10^-34)(3*10^8)/(2.11*10^-18

WL = 9.4208*10^-8 m

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