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 determine the minimum potential difference (V) required to observe the first line of the Balmer series, we start with the energy levels of a hydrogen atom.

Step 1: Write the energy level formula
The energy of an electron in the \(n^{th}\) orbit is given by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]

Step 2: Identify the transition for the first Balmer line
The Balmer series corresponds to transitions ending at \(n_f = 2\) and starting from \(n_i \geq 3\). For the first Balmer line: \[ n_i = 3, \quad n_f = 2 \]

Step 3: Calculate the energy difference
\[ \Delta E_{3 \to 2} = E_2 - E_3 \] Substitute values: \[ E_2 = -\frac{13.6}{2^2} = -3.4 \, \text{eV}, \quad E_3 = -\frac{13.6}{3^2} = -1.51 \, \text{eV} \] \[ \Delta E_{3 \to 2} = (-3.4) - (-1.51) = -1.89 \, \text{eV} \] Since the energy difference represents emitted or absorbed energy, take the magnitude: \[ \Delta E_{3 \to 2} = 1.89 \, \text{eV} \]

Step 4: Relate energy to potential difference
The energy supplied by the potential difference must equal this excitation energy: \[ eV = 1.89 \, \text{eV} \]

Step 5: Solve for the given potential relationship
If the potential difference is given as \( \frac{\alpha}{10} \, V \), then: \[ e \times \frac{\alpha}{10} = 1.89 \, \text{eV} \] \[ \Rightarrow \alpha = 18.9 \]

Step 6: Final verification
This means the potential difference required for the excitation to occur corresponds to: \[ V = \frac{\alpha}{10} = \frac{18.9}{10} = 1.89 \, \text{V} \]

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

The minimum potential difference needed to produce visible Balmer series lines in hydrogen is therefore equivalent to an excitation energy of 1.89 eV, corresponding to \( \alpha = 18.9 \).