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 \( \alpha = \, \underline{\hspace{2cm}} \).
Correct Answer: 121
Solution and Explanation
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 \( \alpha = 18.9 \).
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