The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

<p>The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:</p> <p>$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$</p> <p>where:</p> <ul> <li>$\lambda$ is the wavelength of the emitted light,</li> <li>$R$ is the Rydberg constant,</li> <li>$n_1$ is the lower energy level, and</li> <li>$n_2$ is the higher energy level.</li> </ul> <p>The Balmer series corresponds to electron transitions where the lower energy level, $n_1$, is 2, and the Lyman series corresponds to transitions where $n_1$ is 1. The shortest wavelength in each series occurs for transitions from the highest possible energy level ($n_2 = \infty$) to the specified lower level ($n_1$).</p> <p><strong>For the Balmer series (shortest wavelength)</strong>:</p> <p>$n_1 = 2, n_2 = \infty$,</p> <p>Putting these values into the Rydberg formula gives:</p> <p>$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$</p> <p>$\frac{1}{\lambda_{B}} = R \left( \frac{1}{4} \right)$</p> <p>$\lambda_{B} = \frac{1}{R \cdot \frac{1}{4}} = \frac{4}{R}$</p> <p><strong>For the Lyman series (shortest wavelength)</strong>:</p> <p>$n_1 = 1, n_2 = \infty$,</p> <p>Putting these values into the Rydberg formula gives:</p> <p>$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$</p> <p>$\frac{1}{\lambda_{L}} = R \cdot 1$</p> <p>$\lambda_{L} = \frac{1}{R}$</p> <p><strong>The ratio of the shortest wavelength of Balmer series ($\lambda_{B}$) to the shortest wavelength of Lyman series ($\lambda_{L}$) is:</strong></p> <p>$$\frac{\lambda_{B}}{\lambda_{L}} = \frac{\frac{4}{R}}{\frac{1}{R}} = \frac{4}{1} = 4:1$$</p> <p>Therefore, the correct answer is <strong>Option D: $4: 1$</strong>.</p>