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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, jinawee Feb 7 at 22:32

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Hint: don't use 5.29 pm, use the constants. –  Kyle Kanos Feb 7 at 17:43

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