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

Step 1: Recall the Balmer-series formula.
For hydrogen, $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ with the lower level $n_1 = 2$ for the Balmer series.
Step 2: Insert the given wavelength.
Given $\lambda = \dfrac{7.2}{R}$, so $\dfrac{1}{\lambda} = \dfrac{R}{7.2}$. Cancelling $R$, $\dfrac{1}{7.2} = \dfrac{1}{4} - \dfrac{1}{n_2^2}$.
Step 3: Simplify the constant.
Since $7.2 = \dfrac{36}{5}$, $\dfrac{1}{7.2} = \dfrac{5}{36}$.
Step 4: Solve for the higher level.
$\dfrac{1}{n_2^2} = \dfrac{1}{4} - \dfrac{5}{36} = \dfrac{9 - 5}{36} = \dfrac{4}{36} = \dfrac{1}{9}$, so $n_2 = 3$.
Step 5: Use the orbit-radius law.
In Bohr's model $r_n \propto n^2$, so $\dfrac{r_{higher}}{r_{lower}} = \dfrac{n_2^2}{n_1^2} = \dfrac{9}{4}$.
Step 6: Conclude.
The ratio of the higher to lower orbit radii is $9:4$, which is option (1).
\[ \boxed{r_{higher} : r_{lower} = 9 : 4} \]