GREPhysics.NET: GR9677 #40

GR9677 #40

Problem

GREPhysics.NET Official Solution

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Quantum Mechanics $\Rightarrow$ }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).

Alternate Solutions

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.

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Comments

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.

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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.

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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.

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astro_allison
2005-12-08 22:50:07

don't you mean $E_{f} = \frac{Z^2 E_{0}}{n_{f}^2}$ ?

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don't you mean $E_{f} = \frac{Z^2 E_{0}}{n_{f}^2}$ ?

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