Bohr's Correspondence Principle

[little neutrino]

Homework Statement

(a) Show that in the Bohr model, the frequency of revo-lution of an electron in its circular orbit around a stationary hydrogen nucleus is f = me⁴/4ε₀²n³h³ (b) In classical physics, the frequency of revolution of the electron is equal to the frequency of the radiation that it emits. Show that when n is very large, the fre-quency of revolution does indeed equal the radiated frequency cal-culated from Eq. (39.5) for a transition from n₁ = n + 1 to n₂ = n.

Homework Equations

v = e²/2ε₀nh
r = ε₀n²h²/πme²

The Attempt at a Solution

I managed to solve part (a). But for part (b), I'm not sure how to find the energy of the photon. I tried
E = -13.6eV (1/n² - 1/(n+1)²) which I expanded to get
E = -13.6eV ((2n+1)/n²(n+1)²) but doesn't this tend to 0 as n approaches infinity? Since E = hf this implies that f tends to 0 as well? Does anybody know how to prove the relationship in part (b)? Thanks! :)

 

[blue_leaf77]

but doesn't this tend to 0 as n approaches infinity?

That's the asymptotic value of $\Delta E$, however the problem asks you to calculate the behavior of this quantity when $n$ is very large, therefore it must still contain $n$ in its expression. Consider incorporating the inequality $n\gg 1$ in the last equation.

 

[little neutrino]

That's the asymptotic value of $\Delta E$, however the problem asks you to calculate the behavior of this quantity when $n$ is very large, therefore it must still contain $n$ in its expression. Consider incorporating the inequality $n\gg 1$ in the last equation.

Hmm, if n >> 1, then the numerator 2n + 1 = 2n and the denominator n²(n+1)² = n⁴. After simplifying this, I get the answer. But I'm still a bit confused; if I directly substitute n+1 = n (since n>>1) for the initial expression 1/n² - 1/(n+1)², won't I just get 0? :/

 

[blue_leaf77]

Hmm, if n >> 1, then the numerator 2n + 1 = 2n and the denominator n²(n+1)² = n⁴. After simplifying this, I get the answer. But I'm still a bit confused; if I directly substitute n+1 = n (since n>>1) for the initial expression 1/n² - 1/(n+1)², won't I just get 0? :/

The proper way to go about this problem is actually by employing Taylor expansion. The exact equation of $\Delta E$ can be rewritten as
$$
\Delta E = \frac{13.6}{n^2}\left(1-\frac{1}{(1+1/n)^2}\right)
$$
Now since $n\gg 1$, the second term in the bracket can be expanded into power series (or more accurately, binomial series),
$$
(1+1/n)^{-2} = 1-\frac{2}{n}+\frac{3}{n^2}-\ldots
$$
Truncating this series up to the second term and substituting it into $\Delta E$ will give you the same answer.

 

[little neutrino]

The proper way to go about this problem is actually by employing Taylor expansion. The exact equation of $\Delta E$ can be rewritten as
$$
\Delta E = \frac{13.6}{n^2}\left(1-\frac{1}{(1+1/n)^2}\right)
$$
Now since $n\gg 1$, the second term in the bracket can be expanded into power series (or more accurately, binomial series),
$$
(1+1/n)^{-2} = 1-\frac{2}{n}+\frac{3}{n^2}-\ldots
$$
Truncating this series up to the second term and substituting it into $\Delta E$ will give you the same answer.

Ohhh I see. Thanks so much! :)

 

[epenguin]

I can't tell you anything about Eq. (39.5) as you didn't tell us what it is! But if it is just showing that you go to an inverse cubic relation for high n, then you have nearly done it in the OP - that expression does tend to inverse cube as n gets high (not needing anything so advanced is a Taylor series).

And for your question doesn't f tend to 0, well it's Δf not f - and tending to 0 is exactly what it does do and is seen in the spectra - the lines get closer and closer together. Σ Δf is convergent, so there is a definite limit (corresponding to the escape velocity in classical physics) and the higher spectral lines crowd towards this.