Question

If the wavelength of a spectral line in the Balmer series of hydrogen spectrum is $\frac{7.2}{R}$, then the ratio of the radii of the higher and lower orbits between which the transition of electron takes place is ($R$ - Rydberg constant):

• A. $9:4$
• B. $4:1$
• C. $25:4$
• D. $8:1$

Hint:

For Balmer series questions, always take the lower energy level as $n_1=2$. Once the higher level $n_2$ is found using the Rydberg formula, use Bohr's relation $r_n \propto n^2$ to obtain orbit-radius ratios directly.

Answer 1

The Correct Option is A

Concept: According to the Rydberg formula, the wavelength of a spectral line emitted during an electronic transition in a hydrogen atom is given by \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), \] where:

• $R$ is the Rydberg constant,

• $n_1$ is the lower energy level,

• $n_2$ is the higher energy level.
For the Balmer series, \[ n_1 = 2. \] Also, according to Bohr's model of the hydrogen atom, the radius of the $n^{\text{th}}$ orbit is \[ r_n \propto n^2. \] Therefore, \[ \frac{r_{\text{higher}}}{r_{\text{lower}}} = \frac{n_2^2}{n_1^2}. \] Hence, we first determine the value of the higher orbit $n_2$.

Step 1: Substitute the given wavelength into the Rydberg formula Given, \[ \lambda=\frac{7.2}{R}. \] Therefore, \[ \frac{1}{\lambda} = \frac{R}{7.2}. \] Using the Balmer series relation, \[ \frac{R}{7.2} = R \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right). \] Cancelling $R$ from both sides, \[ \frac{1}{7.2} = \frac{1}{4} - \frac{1}{n_2^2}. \] Since \[ 7.2=\frac{36}{5}, \] we obtain \[ \frac{1}{7.2} = \frac{5}{36}. \] Thus, \[ \frac{5}{36} = \frac{1}{4} - \frac{1}{n_2^2}. \]

Step 2: Calculate the higher energy level Rearranging, \[ \frac{1}{n_2^2} = \frac{1}{4} - \frac{5}{36}. \] Taking LCM $36$, \[ \frac{1}{n_2^2} = \frac{9-5}{36} = \frac{4}{36} = \frac{1}{9}. \] Hence, \[ n_2^2=9 \] and therefore, \[ n_2=3. \] So the electronic transition occurs from \[ n_2=3 \quad \text{to} \quad n_1=2. \]

Step 3: Find the ratio of orbital radii Since \[ r_n \propto n^2, \] we have \[ \frac{r_{\text{higher}}}{r_{\text{lower}}} = \frac{r_3}{r_2} = \frac{3^2}{2^2} = \frac{9}{4}. \] Therefore, \[ r_{\text{higher}} : r_{\text{lower}} = 9 : 4. \] \[ \boxed{9:4} \]