quantum calculations of hydrogen spectrum

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HYDROGEN’S SPECTRAL TUBE
PROMINENT LINES OF EMISSION
What color goes with each electron transition in the visible spectrum?
BLUE: 52 TEAL 42 RED 32
It was experimentally determined that:
En = -13.6 eV
n2
where En is the energy of an electron measured in electron volts
(an energy unit) and n is the principal quantum number
What is the energy of an electron on principal energy level n=5?
E5 = -13.6 eV = -0.544 eV
52
What is the energy of an electron on principal energy level n=2?
E2 = -13.6 eV = -3.4 eV
22
1
If 1 eV=1.6 x 10-19 J (Joules which is another energy unit also
involved in Planck’s constant), what are the values for E5 and E2
of hydrogen?
E5 = -0.544 eV x 1.6 x 10-19 J = -8.704 x 10-20 J
1 eV
E2 = -3.4 eV x 1.6 x 10-19 J = -5.44 x 10-19 J
1 eV
What is the change in energy from E5 down to E2 ?
(triangle is the Greek capital letter delta which means change)
ΔE5 to 2 = Efinal – Einitial = E2 – E5
= -5.44 x 10-19 J – (-8.704 x 10-20 J) = -4.5696 x 10-19 J
=-4.57 x 10-19 J
Why is the value of this transition “negative energy?”
Because the energy is being emitted as the electron falls from
n=5 to n=2.
2
What is the wavelength of the photon of light emitted from
E5 down to E2 ?
IF:
Equation 1 Equantum = h ν
energy = Planck’s constant x frequency
(6.626 x 10-34 J.s)
AND:
Equation 2 c = λ ν
speed of light = wavelength x frequency
(3.00 x 108 m/s)
AND:
Equation 3 ν = _c_
λ
when equation 2 above is rearranged and solved for ν
AND:
Equation 4 Equantum = h c
λ
when ν in equation 1 is substituted by equation 3
THEN:
Equation 5 λ = __h c__
Equantum
when equation 4 is solved for λ
λ5 to 2 = __h c__
E5 to 2
λ = (6.626 x 10-34 J.s)x (3.00 x 108 m/s) = -4.349671772 x 10-7
-4.57 x 10-19 J
= -4.35 x 10-7 m
Unit Cancelers
3
If 1.0 m =1.0 x 109 nm=1.0 x 1010 angstroms (Ǻ is another length
value often used with atoms), what is the wavelength for the E5
and E2 transition of hydrogen in Ǻ and nm?
λ = -4.35 x 10-7 m x 1.0 x 1010 Ǻ = -4.35 x 103 Ǻ = -4350 Ǻ
1.0 m
λ = -4.35 x 10-7 m x 1.0 x 109 nm = -4.35 x 102 nm = -435 nm
1.0 m
Unit Cancelers
ACTUAL VALUE OF HYRDOGEN’S E5 to E2 TRANSITION
λ5 to 2 = 434.17 nm
Does this make sense?
4
HOMEWORK:
1) Show that the E4 to E2 transition should emit photons of
about λ4 to 2 = 486.27 nm
2) Show that the E3 to E2 transition should emit photons of
about λ3 to 2 = 656.47 nm
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