Question

Ionisation potential of hydrogen atom is \(13.6 \, \text{eV}\). Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy \(12.1 \, \text{eV}\). The spectral lines emitted by hydrogen atoms according to Bohr's theory will be:

• A. one
• B. two
• C. three
• D. four

Hint:

If electron reaches level \(n\), total spectral lines = \(\frac{n(n-1)}{2}\).

Answer 1

The Correct Option is C

Concept: Energy levels of hydrogen: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]

Step 1: Ground state energy \[ E_1 = -13.6 \, \text{eV} \]

Step 2: Check excitation level Given photon energy \(= 12.1 \, \text{eV}\) \[ E_3 = -\frac{13.6}{9} = -1.51 \, \text{eV} \] \[ \Delta E = E_3 - E_1 = (-1.51) - (-13.6) = 12.09 \approx 12.1 \, \text{eV} \] \[ \Rightarrow \text{electron is excited to } n=3 \]

Step 3: Possible transitions From \(n=3\), electron can undergo: \[ 3 \rightarrow 2,\quad 3 \rightarrow 1,\quad 2 \rightarrow 1 \]

Step 4: Number of spectral lines \[ \text{Total lines} = \frac{n(n-1)}{2} = \frac{3 \cdot 2}{2} = 3 \] Conclusion \[ \text{Number of spectral lines} = 3 \]