Question

A hydrogen atom is bombarded with electrons accelerated through a potential difference of \( V \), which causes excitation of hydrogen atoms. If the experiment is being performed at \( T = 0 \, \text{K} \), the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be\[\frac{\alpha}{10} V,\]where \( \alpha = \, \underline{\hspace{2cm}} \).

Answer 1

Correct Answer: 121

To ascertain the minimum potential difference (V) necessary for observing the first line of the Balmer series, we begin with the energy levels of a hydrogen atom.

Step 1: Energy Level Formula
The energy of an electron in the \(n^{th}\) orbit is expressed as: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]

Step 2: First Balmer Line Transition Identification
The Balmer series involves transitions terminating at \(n_f = 2\) and originating from \(n_i \geq 3\). For the initial Balmer line: \[ n_i = 3, \quad n_f = 2 \]

Step 3: Energy Difference Calculation
\[ \Delta E_{3 \to 2} = E_2 - E_3 \] Substituting values: \[ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV}, \quad E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} \] \[ \Delta E_{3 \to 2} = (-3.4) - (-1.51) = -1.89 \, \text{eV} \] Considering the energy difference as either emitted or absorbed energy, we take its magnitude: \[ \Delta E_{3 \to 2} = 1.89 \, \text{eV} \]

Step 4: Energy and Potential Difference Relationship
The energy provided by the potential difference must equate to this excitation energy: \[ e V = 1.89 \, \text{eV} \]

Step 5: Solving for the Given Potential Relationship
Given the potential difference as \( \frac{\alpha}{10} \, V \), we have: \[ e \times \frac{\alpha}{10} = 1.89 \, \text{eV} \] \[ \Rightarrow \alpha = 18.9 \]

Step 6: Final Verification
This indicates that the potential difference required for the excitation is: \[ V = \frac{\alpha}{10} = \frac{18.9}{10} = 1.89 \, \text{V} \]

Final Answer:
\[ \boxed{\alpha = 18.9} \]

Consequently, the minimum potential difference required to generate visible Balmer series lines in hydrogen is equivalent to an excitation energy of 1.89 eV, corresponding to \( \alpha = 18.9 \).