Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is

Options

A. `10.20 eV`
B. `20.40 eV`
C. `13.6 eV`
D. `27.2 eV`

Explanation

To solve the problem of the maximum reduction in kinetic energy when two hydrogen atoms collide inelastically, we can follow these steps:

### Step 1: Determine the initial kinetic energy of the hydrogen atoms
Each hydrogen atom in the ground state has a kinetic energy of 13.6 eV. Therefore, for two hydrogen atoms, the total initial kinetic energy (KE_initial) is:
\[ KE_{\text{initial}} = 2 \times 13.6 \, \text{eV} = 27.2 \, \text{eV} \]

### Step 2: Analyze the inelastic collision
In an inelastic collision, momentum is conserved, but kinetic energy is not. After the collision, we can assume that one hydrogen atom remains in the ground state (n=1) and the other transitions to the first excited state (n=2).

### Step 3: Calculate the kinetic energy after the collision
The energy of a hydrogen atom in a given state is given by the formula:
\[ E_n = \frac{13.6 \, \text{eV}}{n^2} \]
For the ground state (n=1):
\[ E_1 = \frac{13.6 \, \text{eV}}{1^2} = 13.6 \, \text{eV} \]
For the first excited state (n=2):
\[ E_2 = \frac{13.6 \, \text{eV}}{2^2} = \frac{13.6 \, \text{eV}}{4} = 3.4 \, \text{eV} \]

Now, the total kinetic energy after the collision (KE_final) is:
\[ KE_{\text{final}} = E_1 + E_2 = 13.6 \, \text{eV} + 3.4 \, \text{eV} = 17.0 \, \text{eV} \]

### Step 4: Calculate the reduction in kinetic energy
The reduction in kinetic energy (ΔKE) is given by the difference between the initial and final kinetic energies:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 27.2 \, \text{eV} - 17.0 \, \text{eV} = 10.2 \, \text{eV} \]

### Step 5: Conclusion
The maximum amount by which their combined kinetic energy is reduced is:
\[ \Delta KE = 10.2 \, \text{eV} \]