Question

The number of de-Broglie wavelengths contained in the second Bohr orbit of hydrogen atom is

Answer 1

2

Answer 2

Explanation of the solution:

According to Bohr's quantization condition, the angular momentum of an electron in a stable orbit is quantized, given by $mvr_n = n\frac{h}{2\pi}$, where $n$ is the principal quantum number. According to de-Broglie's hypothesis, the wavelength of the electron is $\lambda = \frac{h}{mv}$. Combining these two equations, we get $mv = \frac{h}{\lambda}$. Substituting this into the Bohr condition:

$(\frac{h}{\lambda})r_n = n\frac{h}{2\pi}$

$\frac{r_n}{\lambda} = \frac{n}{2\pi}$

$2\pi r_n = n\lambda$

This equation shows that the circumference of the $n$-th Bohr orbit ($2\pi r_n$) is equal to $n$ times the de-Broglie wavelength ($\lambda$) of the electron in that orbit. This means that exactly $n$ de-Broglie wavelengths fit into the circumference of the $n$-th orbit, forming a standing wave. The number of de-Broglie wavelengths contained in the $n$-th orbit is therefore $n$.

For the second Bohr orbit, the principal quantum number is $n=2$.

Thus, the number of de-Broglie wavelengths contained in the second Bohr orbit is 2.