To solve the problem, we need to find the kinetic energy of the colliding electron that results in the excitation of a hydrogen atom from its ground state, leading to the emission of a photon corresponding to the largest wavelength of the Balmer series. Let's break down the solution step by step:
### Step 1: Understand the Energy Levels of Hydrogen
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.
### Step 2: Determine the Energy of the Ground State
For the ground state (n=1):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
### Step 3: Identify the Transition for the Largest Wavelength in the Balmer Series
The largest wavelength in the Balmer series corresponds to the transition from \( n=3 \) to \( n=2 \). We need to calculate the energy difference between these two levels.
### Step 4: Calculate the Energy of the Excited State (n=3)
For \( n=3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]
### Step 5: Calculate the Energy Difference for the Transition
The energy difference \( \Delta E \) for the transition from \( n=3 \) to \( n=2 \) is given by:
\[ \Delta E = E_2 - E_3 \]
Where:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
Now, substituting the values:
\[ \Delta E = (-3.4 \, \text{eV}) - (-1.51 \, \text{eV}) = -3.4 + 1.51 = -1.89 \, \text{eV} \]
Since we are interested in the absolute value:
\[ \Delta E = 1.89 \, \text{eV} \]
### Step 6: Relate the Energy of the Photon to the Kinetic Energy of the Electron
Since the electron loses all its kinetic energy to excite the hydrogen atom, the kinetic energy of the colliding electron will be equal to the energy of the emitted photon:
\[ KE = \Delta E \]
### Step 7: Conclusion
Thus, the kinetic energy of the colliding electron is approximately:
\[ KE \approx 1.9 \, \text{eV} \]
### Final Answer
The kinetic energy of the colliding electron will be **1.9 electron volts**.
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