An electron collides with a hydrogen atom in its ground state and excites it to `n = 3` ,. The energy gives to hydrogen atom n this inelastic collision is [Neglect the recoiling of hydrogen atom]

Options

A. `10.2 eV`
B. `12.1 eV`
C. `12.5 eV`
D. none of these

Explanation

To solve the problem of how much energy is given to a hydrogen atom in its ground state when it is excited to n = 3 by an electron collision, 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.

### Step 2: Calculate the Energy of the Ground State (n = 1)
For the ground state (n = 1):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

### Step 3: Calculate the Energy of the Excited State (n = 3)
For the excited state (n = 3):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \]
Calculating this gives:
\[ E_3 = -1.51 \, \text{eV} \]

### Step 4: Calculate the Energy Difference
The energy given to the hydrogen atom during the collision is the difference between the energy of the excited state and the ground state:
\[ \Delta E = E_3 - E_1 \]
Substituting the values:
\[ \Delta E = (-1.51 \, \text{eV}) - (-13.6 \, \text{eV}) \]
\[ \Delta E = -1.51 + 13.6 = 12.09 \, \text{eV} \]

### Step 5: Final Answer
The energy given to the hydrogen atom in this inelastic collision is approximately:
\[ \Delta E \approx 12.1 \, \text{eV} \]