Question

A free electron of 2.6 eV energy collides with a \(H^+\) ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (\(h = 6.6 \times 10^{-34}\) J s)

• A. \(0.19 \times 10^{15}\) MHz
• B. \(9.0 \times 10^{27}\) MHz
• C. \(1.45 \times 10^9\) MHz
• D. \(1.45 \times 10^{16}\) MHz

Hint:

Remember: Ground state is \(n=1\), First Excited State is \(n=2\), Second Excited State is \(n=3\). Always convert eV to Joules (\(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\)) before calculating frequency in SI units.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a free electron is captured by an ion to form an atom in a bound state, the total energy must be conserved. The energy of the emitted photon is equal to the difference between the initial total energy of the free system and the final energy of the bound hydrogen atom.
Step 2: Key Formula or Approach:
1. Initial Energy (\(E_i\)) = Kinetic Energy of free electron + Potential energy at infinity (assumed zero).
2. Final Energy (\(E_f\)) = Energy of the H-atom in the \(n^{th}\) state = \(-\frac{13.6}{n^2}\) eV.
3. Photon Energy (\(E_{ph}\)) = \(E_i - E_f = h\nu\).
Step 3: Detailed Explanation:
Given:
Initial energy of electron, \(E_i = 2.6\) eV.
The atom is formed in the first excited state, which means \(n = 2\).
Energy of hydrogen atom in \(n=2\):
\[ E_f = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV} \]
Energy released as a photon:
\[ E_{ph} = E_i - E_f = 2.6 - (-3.4) = 2.6 + 3.4 = 6.0 \text{ eV} \]
Convert this energy to Joules:
\[ E_{ph} = 6.0 \times 1.6 \times 10^{-19} \text{ J} = 9.6 \times 10^{-19} \text{ J} \]
Frequency (\(\nu\)) calculation:
\[ \nu = \frac{E_{ph}}{h} = \frac{9.6 \times 10^{-19}}{6.6 \times 10^{-34}} \]
\[ \nu \approx 1.4545 \times 10^{15} \text{ Hz} \]
To convert to MHz (\(10^6\) Hz):
\[ \nu = \frac{1.4545 \times 10^{15}}{10^6} \text{ MHz} = 1.4545 \times 10^9 \text{ MHz} \]
Step 4: Final Answer:
The frequency of the emitted photon is approximately \(1.45 \times 10^9\) MHz.