Question

The de Broglie wavelength of an electron in the first Bohr orbit is

Answer 1

Answer on Question #78635, Physics / Atomic and Nuclear Physics

The de Broglie wavelength of an electron in the first Bohr orbit is

Solution:

According to the Bohr model of the atom (https://en.wikipedia.org/wiki/Bohr_model) the angular momentum of the orbiting electron is quantised such that

$$m v r = \frac {n h}{2 \pi}.$$

From this condition the de Broglie wavelength of an electron in the first Bohr orbit $(n = 1)$ is

$$\lambda = \frac {h}{m v} = 2 \pi r = 0.332 \, \text{nm}.$$

Note, herer $= 0.0529 \, \text{nm}$ being the Bohr radius.

**Answer:** $\lambda = 0.332 \, \text{nm}$.

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