The highest excited state that an unexcited hydrogen atom can reach when they are bombarded with ` 12.2 eV` electron is :

Options

A. n=1
B. n=2
C. n=3
D. n=4

Answer

To determine the highest excited state that an unexcited hydrogen atom can reach when bombarded with 12.2 eV of energy, we can follow these steps:

### 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 (1, 2, 3, ...).

### Step 2: Identify the Ground State Energy
For a hydrogen atom in the ground state (n=1), the energy is:
\[ E_1 = -13.6 \, \text{eV} \]

### Step 3: Calculate Total Energy After Bombardment
When the hydrogen atom is bombarded with 12.2 eV of energy, the total energy of the atom becomes:
\[ E_{\text{total}} = E_1 + 12.2 \, \text{eV} = -13.6 \, \text{eV} + 12.2 \, \text{eV} = -1.4 \, \text{eV} \]

### Step 4: Set Up the Equation for the Excited State Energy
We need to find the principal quantum number \( n \) for which the energy \( E_n \) equals -1.4 eV:
\[ -\frac{13.6 \, \text{eV}}{n^2} = -1.4 \, \text{eV} \]

### Step 5: Solve for \( n^2 \)
Rearranging the equation gives:
\[ \frac{13.6}{n^2} = 1.4 \]
\[ n^2 = \frac{13.6}{1.4} \approx 9.7 \]

### Step 6: Calculate \( n \)
Taking the square root of both sides:
\[ n \approx \sqrt{9.7} \approx 3.1 \]
Since \( n \) must be a whole number, we round down to the nearest whole number, which is 3.

### Conclusion
The highest excited state that the hydrogen atom can reach when bombarded with 12.2 eV of energy is \( n = 3 \).

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