A hydrogen atom `(mass = 1.66 xx 10^(-27)` kg) ionzation potential = 13.6eV), moving with a velocity `(6.24 xx 10^(4) m s^(-1)` makes a completely inelastic head-on collision with another stationary hydrogen atom. Both atoms are in the ground state before collision . Up to what state either one atom may be excited?

Explanation

To solve the problem of how much a hydrogen atom can be excited after a completely inelastic collision with another stationary hydrogen atom, we can follow these steps:

### Step 1: Calculate the initial kinetic energy of the moving hydrogen atom.
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m = 1.66 \times 10^{-27} \) kg (mass of hydrogen atom)
- \( v = 6.24 \times 10^{4} \) m/s (velocity of the moving hydrogen atom)

Substituting the values:
\[ KE = \frac{1}{2} \times (1.66 \times 10^{-27}) \times (6.24 \times 10^{4})^2 \]

### Step 2: Calculate the final velocity after the collision.
In a completely inelastic collision, the two atoms stick together and move with a common velocity. The initial momentum before the collision is equal to the final momentum after the collision.

Initial momentum:
\[ p_{initial} = m \cdot v = (1.66 \times 10^{-27}) \cdot (6.24 \times 10^{4}) \]

Final momentum (both atoms moving together):
\[ p_{final} = 2m \cdot v_{final} \]

Setting initial momentum equal to final momentum:
\[ (1.66 \times 10^{-27}) \cdot (6.24 \times 10^{4}) = 2 \cdot (1.66 \times 10^{-27}) \cdot v_{final} \]

Solving for \( v_{final} \):
\[ v_{final} = \frac{(1.66 \times 10^{-27}) \cdot (6.24 \times 10^{4})}{2 \cdot (1.66 \times 10^{-27})} = \frac{6.24 \times 10^{4}}{2} = 3.12 \times 10^{4} \text{ m/s} \]

### Step 3: Calculate the loss of kinetic energy due to the collision.
The loss of kinetic energy (\( \Delta KE \)) can be calculated as:
\[ \Delta KE = KE_{initial} - KE_{final} \]
Where:
\[ KE_{final} = \frac{1}{2} \cdot 2m \cdot v_{final}^2 = m \cdot v_{final}^2 \]

Substituting the values:
\[ KE_{final} = (1.66 \times 10^{-27}) \cdot (3.12 \times 10^{4})^2 \]

Now, calculate \( \Delta KE \):
\[ \Delta KE = KE_{initial} - KE_{final} \]

### Step 4: Convert the loss of kinetic energy to electron volts.
To convert joules to electron volts, use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]

So, the energy in eV is:
\[ \Delta KE_{eV} = \frac{\Delta KE}{1.6 \times 10^{-19}} \]

### Step 5: Determine the maximum excitation state.
The energy required to excite a hydrogen atom from the ground state (n=1) to the nth state is given by:
\[ E_n = 13.6 \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \text{ eV} \]

Setting \( \Delta KE_{eV} \) equal to \( E_n \):
\[ \Delta KE_{eV} = 13.6 \left( 1 - \frac{1}{n^2} \right) \]

Solving for \( n \):
\[ \frac{\Delta KE_{eV}}{13.6} = 1 - \frac{1}{n^2} \]
\[ \frac{1}{n^2} = 1 - \frac{\Delta KE_{eV}}{13.6} \]
\[ n^2 = \frac{1}{1 - \frac{\Delta KE_{eV}}{13.6}} \]

Finally, calculate \( n \).