Rankers Physics

Atomic Structure: Practice Problem & Solution

Given the value of Rydberg constant is $10^7 m^{-1}$, the wave number of the last line of the Balmer series in hydrogen spectrum will be: (2016 - I)
$0.025 \times 10^4 m^{-1}$
$0.5 \times 10^7 m^{-1}$
$0.25 \times 10^7 m^{-1}$
$2.5 \times 10^7 m^{-1}$

Solution Explained:

To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:

The wave number $\bar{\nu}$ is $1/\lambda$. For the last line of the Balmer series, $n_1 = 2$ and $n_2 = \infty$. Thus, $\bar{\nu} = R(\frac{1}{2^2} - 0) = \frac{R}{4} = \frac{10^7}{4} = 0.25 \times 10^7 m^{-1}$.

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