Question

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A∘ . The wavelength of the second member of the Balmer series (in nm) is_____________

Answer 1

486.0 nm

Answer 2

The Balmer series corresponds to electron transitions to the $n_1=2$ energy level in a hydrogen atom. The first member of the Balmer series results from the transition from $n_2=3$ to $n_1=2$. The second member results from the transition from $n_2=4$ to $n_1=2$. The Rydberg formula for the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ For the first member ($\lambda_1$, $n_1=2$, $n_2=3$): $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36}$ We are given $\lambda_1 = 6561 \text{ A}^\circ$.

For the second member ($\lambda_2$, $n_1=2$, $n_2=4$): $\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{16-4}{64} \right) = \frac{12R}{64} = \frac{3R}{16}$ Now, we can find the ratio of the wavelengths: $\frac{1/\lambda_2}{1/\lambda_1} = \frac{3R/16}{5R/36}$ $\frac{\lambda_1}{\lambda_2} = \frac{3}{16} \times \frac{36}{5} = \frac{3 \times 9}{4 \times 5} = \frac{27}{20}$ So, $\lambda_2 = \lambda_1 \times \frac{20}{27}$. Given $\lambda_1 = 6561 \text{ A}^\circ$: $\lambda_2 = 6561 \text{ A}^\circ \times \frac{20}{27} = 243 \times 20 \text{ A}^\circ = 4860 \text{ A}^\circ$ To convert to nanometers (nm), we use the conversion $1 \text{ A}^\circ = 0.1 \text{ nm}$: $\lambda_2 = 4860 \text{ A}^\circ \times 0.1 \text{ nm/A}^\circ = 486.0 \text{ nm}$