If a singly ionized Helium atom in an n = 4 state emits a photon of wavelength 470 nanometers, which of the following gives the approximate final energy level Ef of the atom, and the n value, of nf this final state?
Ef (eV)
|
nf
| |
| A. |
− 6.0
|
3
|
| B. |
− 6.0
|
2
|
| C. |
− 14
|
2
|
| D. |
− 14
|
1
|
| E. |
− 52
|
1
|
(GR9677 #40)
Ephoton = Ei − Ef
Helium: 2 electrons, 2 protons, 2 neutron
Singly ionized Helium, He+:1 electrons, 2 protons, 2 neutron
He+ → Hydrogen-like atom
Bohr's Equation for Hydrogen-like atom: En = −13.6 Z2/n2 eV
For Helium, Z = 2,
Ei = E(n =4) = − 13.6 (2)2 /42 = − 13.6 /4 ≈ − 3.4 eV
Ephoton = hν = hc / λ
with
h = 6.63 × 10−34 Joule.second = 4.1 × 10−15 eV.second
c = 3 × 108 m/s
λ = 470 nm = 470 × 10−9 m = 4.7 × 10−7 m
Ephoton = (4.1 × 10−15)(3 × 108) / (4.7 × 10−7) ≈ 3 eV
Ef = Ei − Ephoton
= − 3.4 − 3
= − 6.4 eV
To find nf :
nf 2 = −13.6 Z2/ Ef
= −13.6 (2)2/ (− 6.4)
= 54.4/6.4 ≈ 9
nf = 3
Answer: A
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