# Show that the circumference of the Bohr orbit of hydrogen is an integral multiple of the wavelength of the electron in it [closed]

We have an electron revolving in a the $n$th Bohr orbit of an atom of hydrogen. The problem is to prove that the circumference of the orbit is an integral multiple of wavelength of this electron $\lambda_e$.

$$2\pi r_n = k\lambda_e, \ k \in \mathbb{N}$$

where $r_n$ is the radius of the $n$th Bohr orbit. I started out by figuring out by expanding out the different terms in the equation:

The $n$th Bohr radius for hydrogen is given by:

$$r_n = 5.29n^2 \rm{pm}$$

$$\implies 2\pi r_n = 10.58n^2\pi\tag1$$

The wavelength of an electron is given by:

$$\lambda_e = \frac{h}{p_e}$$

The angular momentum of an electron is given by:

$$p_e = k_1\hbar, \ k_1 \in \mathbb{N}$$

$$\implies \lambda_e = \frac{h}{k_1\hbar} = \frac{2\pi}{k_1}\tag2$$

How do I show that $(1)$ is an integral multiple of $(2)$?

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## closed as off-topic by Chris White, Emilio Pisanty, Kyle Kanos, Brandon Enright, jinaweeFeb 7 '14 at 22:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Chris White, Emilio Pisanty, Kyle Kanos, Brandon Enright, jinawee
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Hint: don't use 5.29 pm, use the constants. – Kyle Kanos Feb 7 '14 at 17:43