Question

A series of lines in the spectrum of atomic H lies at wavelengths 656.46486.27, 434.17, 410.29 mm . What is the wavelength of next line in this series?

• A. 356
• B. 375
• C. 397.2
• D. 328

Answer 1

Answer: (C)

397.2

Answer 2

The given wavelengths (656.46 nm, 486.27 nm, 434.17 nm, 410.29 nm) correspond to the Balmer series of the hydrogen atom, where transitions end at the energy level $n_f = 2$. These lines are from transitions originating from higher energy levels $n_i = 3, 4, 5, 6$ respectively. The Rydberg formula for the wavelength $\lambda$ of spectral lines in a hydrogen atom is: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ where $R_H$ is the Rydberg constant.

For the Balmer series ($n_f=2$), the wavelengths are for $n_i=3, 4, 5, 6$. The question asks for the wavelength of the next line in this series, which corresponds to the transition from the next higher energy level, $n_i=7$, to $n_f=2$.

Using the first line $\lambda_1 = 656.46 \, \text{nm}$ for $n_i=3$ and $n_f=2$, we can determine the effective Rydberg constant for this problem: $R_H = \frac{1}{\lambda_1} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)^{-1} = \frac{1}{656.46 \, \text{nm}} \left( \frac{1}{2^2} - \frac{1}{3^2} \right)^{-1}$ $R_H = \frac{1}{656.46 \, \text{nm}} \left( \frac{1}{4} - \frac{1}{9} \right)^{-1} = \frac{1}{656.46 \, \text{nm}} \left( \frac{9-4}{36} \right)^{-1} = \frac{1}{656.46 \, \text{nm}} \left( \frac{5}{36} \right)^{-1}$ $R_H = \frac{36}{5 \times 656.46} \, \text{nm}^{-1} = \frac{7.2}{656.46} \, \text{nm}^{-1} \approx 0.0109678 \, \text{nm}^{-1}$

Now, we calculate the wavelength for the next line, corresponding to the transition from $n_i=7$ to $n_f=2$: $\frac{1}{\lambda_{\text{next}}} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) = R_H \left( \frac{1}{2^2} - \frac{1}{7^2} \right)$ $\frac{1}{\lambda_{\text{next}}} = R_H \left( \frac{1}{4} - \frac{1}{49} \right) = R_H \left( \frac{49-4}{196} \right) = R_H \left( \frac{45}{196} \right)$ Substitute the value of $R_H$: $\frac{1}{\lambda_{\text{next}}} = (0.0109678 \, \text{nm}^{-1}) \times \left( \frac{45}{196} \right) \approx 0.0109678 \times 0.2295918 \, \text{nm}^{-1} \approx 0.0025177 \, \text{nm}^{-1}$ $\lambda_{\text{next}} = \frac{1}{0.0025177 \, \text{nm}^{-1}} \approx 397.17 \, \text{nm}$ The calculated wavelength is approximately 397.17 nm. Comparing this with the given options, 397.2 nm is the closest value.

The series limit for the Balmer series is when $n_i \to \infty$, giving $\lambda_{\text{limit}} = \frac{4}{R_H} \approx \frac{4}{0.0109678} \approx 364.7 \, \text{nm}$. Options 356 nm and 328 nm are below this limit and thus are not possible for the Balmer series.