Question

The energy of first (lowest) Balmer line of H atom is \(x\) J. The energy (in J) of second Balmer line of H atom is:

• A. \(\dfrac{x}{1.35}\)
• B. \(x^2\)
• C. \(1.35x\)
• D. \(2x\)

Hint:

Higher Balmer lines have greater transition energy because the initial energy level is higher.

Answer 1

The Correct Option is C

Step 1: Identify transitions for Balmer lines.
Balmer series corresponds to transitions ending at \(n=2\).
First (lowest) Balmer line: \[ n=3 \rightarrow 2 \] Second Balmer line: \[ n=4 \rightarrow 2 \]
Step 2: Use energy expression for hydrogen spectral lines.
\[ E = 13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ eV} \]
Step 3: Write energies of first and second Balmer lines.
First Balmer line: \[ E_1 = 13.6\left(\frac{1}{2^2}-\frac{1}{3^2}\right) = 13.6\left(\frac{5}{36}\right) \] Second Balmer line: \[ E_2 = 13.6\left(\frac{1}{2^2}-\frac{1}{4^2}\right) = 13.6\left(\frac{3}{16}\right) \]
Step 4: Find ratio of energies.
\[ \frac{E_2}{E_1} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{27}{20} \approx 1.35 \] \[ E_2 = 1.35\,E_1 = 1.35x \]
Final Answer: \[ \boxed{1.35x} \]