Question

1. a. A photon is absorbed by a hydrogen atom causing an electron to become excited (nf = 6) from the ground state electron configuration. What is the energy change of the electron associated with this transition?

b. After some time in the excited state, the electron falls from the n = 6 state back to its ground state. What is the change in energy of the electron associated with this transition?

c. When the electron returns from its excited state to its ground state, a photon is emitted. Calculate the wavelength (in nanometers) and the frequency of this photon.

Answer #1

Q1

a)

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)*(1/6^2 – 1/1 ^2)

E = 2.1175*10^-18 J/photon

b)

gaineddE = -Elost

then

E = -2.1175*10^-18J/photon, negative because it is losing it

c)

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)/(2.1175*10^-18)

WL = 9.3874*10^-8 m

to nanometers:

WL = (9.3874*10^-8)(10^9) = 93.87 nm

v = c/WL = (3*10 ^8)/( 9.3874*10^-8) = 3.19*10^15 Hz

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