Question

Line corresponding to lyman series are $\text{L}_1, \text{L}_2, \text{L}_3, \text{L}_4, \ldots$, among these $\text{L}_1$ line corresponds to lowest energy. Similarly lines corresponding to balmer series are $\text{B}_1, \text{B}_2, \text{B}_3, \text{B}_4, \ldots$, among these $\text{B}_1$ line corresponds to lowest energy $\Delta E_L = \text{Energy of } 1^{\text{st}} \text{ line of lyman series}$, $\Delta E_B = \text{Energy of } 1^{\text{st}} \text{ line of balmer series}$. If $\Delta E_L = x \cdot \Delta E_B$. Calculate $(x \times 10^{-1})$

• A. 54
• B. 27
• C. 18
• D. 36

Hint:

The energy of the first line (lowest energy) in the Lyman series ($n=2 \rightarrow 1$) is $3/4$ Rydberg units. The first line in the Balmer series ($n=3 \rightarrow 2$) is $5/36$ Rydberg units.

Answer 1

The Correct Option is A

Energy ratio $\Delta E \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
$L_1$ (Lyman, $n_1=1, n_2=2$): $\Delta E_L \propto \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4}$.
$B_1$ (Balmer, $n_1=2, n_2=3$): $\Delta E_B \propto \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5}{36}$.
$x = \frac{\Delta E_L}{\Delta E_B} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4$.
The question asks for $x \times 10^{-1} = 5.4 \times 0.1 = 0.54$.
However, since the required answer is $54$, we must assume a typo in the question and that it intends to ask for $x \times 10^1$:
$x \times 10^1 = 5.4 \times 10 = 54$.