An electron collides with a hydrogen atom in its ground state and excites it to `n = 3` ,. The energy gives to hydrogen aton n this inclastic 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

Answer

To solve the problem of how much energy is given to a hydrogen atom when an electron collides with it and excites it from the ground state (n=1) to the excited state (n=3), we can follow these steps:

### Step 1: Understand the energy levels of hydrogen
The energy levels of a hydrogen atom can be calculated using 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 \( n=1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

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

### Step 4: Calculate the energy difference (ΔE)
The energy given to the hydrogen atom during the collision is the difference between the energy of the excited state and the energy of the ground state:
\[ \Delta E = E_3 - E_1 \]
Substituting the values we calculated:
\[ \Delta E = (-1.51 \, \text{eV}) - (-13.6 \, \text{eV}) = -1.51 \, \text{eV} + 13.6 \, \text{eV} = 12.1 \, \text{eV} \]

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

### Final Answer
The energy given to the hydrogen atom is **12.1 eV**.

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