1. Using Bohr's second postulate of quantization of orbital angul… - Vidyadip

Question

Using Bohr's second postulate of quantization of orbital angular momentum show that the circumference of the electron in the $n^{th}$ orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.
The electron in hydrogen atom is initially in the third excited state.What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

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Answer

According to Bohr's second postulate

$\text{mvr}_{n} =\frac{\text{nh}}{2\pi}$
$\Rightarrow2\pi\text{r}_{n} = \frac{\text{nh}}{\text{mv}}$
But $\frac{\text{h}}{\text{mv}} =\frac{\text{h}}{\text{p}} =\lambda$
$\therefore2\pi\text{r}_{n} = \text{n}\lambda$

For third excited state n = 4

for ground state n = 1
Hence possible transitions are
$n_1 = 4$ to $n_1 = 3, 2, 1$
$n_1 = 3$ to $n_1 = 2, 1$
$n_1 = 2$ to $n_1 = 1$
Total number of transitions = 6
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