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Quantum Mechanics}Bohr Theory

It's amazing how far one can get with the Bohr formula.

To start with, one should calculate the ground-state energy of the singly ionized Helium (i.e., the ionization energy). E_1 = Z^2 E_{H1} = 4 \times 13.6 eV, since Helium has 2 protons. (The general formula is E_n = Z^2/n^2 E_1.)

The Bohr formula gives E=E_1\left(1/n_f^2-1/n_i^2\right)=E_1(1/n_f^2-1/16), since n_i^2=4^2=16.

E=hc/\lambda\approx 1.24E-6/4.7E-7 gives E\approx 2.5 eV.

The only unknown expression above is n_f. Plugging everything in and solving for that, n_f^{-2}\approx 8^{-1} \Rightarrow n_f\approx 3. This yields choice (A). One can check via E_f=E_1/n_f^2 4\times 13.6/9 \approx 6, which verifies (A).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
BerkeleyEric
2010-09-22 22:36:21		I think process of elimination is fastest here. From the Bohr formula we know that the n=1 is at 4*(-13.6 eV) ~ -52 eV. So it can't be (D). And it also can't be (B) since the n=2 state wouldn't be all the way up at -6 eV.

Now remember that 470 nm falls in the visible regime, so the energy difference must be at around 1-2 eV. This immediately eliminates (C) and (E) since they must be more than this amount lower than the n=4 state.		Alternate Solution - Unverified
archard
2010-07-15 13:50:25		You can also get it with the Rydberg formula, since you're given the wavelength of the photon and the initial energy level. Then it's just plug and chug.Alternate Solution - Unverified
Comments
cczako
2013-10-18 15:21:38		The way I did was to just remember that E is proportional to Z^2/n^2. Its quick and easy to calculate that n=4 --> E=- 3 eV, n=3 --> E=-6 eV, n=2 --> E=-13.6 eV, and n=1 --> E=-50 eV. This automatically eliminates B and D. Then using the equation E=(hc)/lambda (I always use 1240 nm*eV for hc) you get that 470 nm is about 3 eV. So n=4 of energy -3 eV minus 3 eV gives you -6 eV. This gives you choice A.NEC
BerkeleyEric
2010-09-22 22:36:21		I think process of elimination is fastest here. From the Bohr formula we know that the n=1 is at 4*(-13.6 eV) ~ -52 eV. So it can't be (D). And it also can't be (B) since the n=2 state wouldn't be all the way up at -6 eV.

Now remember that 470 nm falls in the visible regime, so the energy difference must be at around 1-2 eV. This immediately eliminates (C) and (E) since they must be more than this amount lower than the n=4 state.		Alternate Solution - Unverified
archard
2010-07-15 13:50:25		You can also get it with the Rydberg formula, since you're given the wavelength of the photon and the initial energy level. Then it's just plug and chug.
sina2
2013-09-24 02:17:02		 I tried in this way, but I failed.
nasim
2015-10-12 17:05:01		 One should first have the Rydberg constant to be able to use Rydberg formula! But doesn\'t!
Alternate Solution - Unverified
asdfman
2009-11-05 00:37:16		You can quickly narrow this down as anything less than ~400 nm is UV. If you remember that 13.6 eV, from hydrogen yields UV then that tosses out choices C, D, and E. NEC
f4hy
2009-10-25 19:28:33		I am confused. When finding E = \frac{hc}{\lambda} are you using the speed of light in cm but the wave length in meters?
kroner
2009-10-29 14:13:19		 Everything is in meters there.
hc \approx 1.24E-6 eV m or 1240 eV nm,
which good value to know off-hand for problems like this.
NEC
astro_allison
2005-12-08 22:50:07		don't you mean E_{f} = \frac{Z^2 E_{0}}{n_{f}^2}?NEC

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The way I did was to just remember that E is proportional to Z^2/n^2. Its quick and easy to calculate that n=4 --> E=- 3 eV, n=3 --> E=-6 eV, n=2 --> E=-13.6 eV, and n=1 --> E=-50 eV. This automatically eliminates B and D. Then using the equation E=(hc)/lambda (I always use 1240 nm*eV for hc) you get that 470 nm is about 3 eV. So n=4 of energy -3 eV minus 3 eV gives you -6 eV. This gives you choice A.

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