A hydrogen atom moving at speed `upsilon` collides with another hydrogen atom kept at rest .Find the minimum value of u for which one of the atoms may get ionized the mass of a hydrogen atom `= 1.67 xx 10^(-27)kg`

Explanation

To find the minimum speed \( u \) at which a moving hydrogen atom can ionize another hydrogen atom at rest, we can follow these steps:

### Step 1: Understand the Problem
We need to find the minimum speed \( u \) of a hydrogen atom that collides with another hydrogen atom at rest to ionize it. The energy required for ionization of a hydrogen atom is given as \( \Delta E = 13.6 \, \text{eV} \).

### Step 2: Convert Ionization Energy to Joules
First, we convert the ionization energy from electron volts to joules:
\[ \Delta E = 13.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.176 \times 10^{-18} \, \text{J} \]

### Step 3: Apply Conservation of Momentum
Using the conservation of momentum, we can express the momentum before and after the collision. Let the mass of the hydrogen atom be \( m = 1.67 \times 10^{-27} \, \text{kg} \).
Before the collision:
\[ \text{Initial momentum} = m \cdot u \]
After the collision, let \( v_1 \) and \( v_2 \) be the velocities of the two hydrogen atoms:
\[ \text{Final momentum} = m \cdot v_1 + m \cdot v_2 \]
Setting these equal gives:
\[ m \cdot u = m \cdot v_1 + m \cdot v_2 \]
Cancelling \( m \) from both sides:
\[ u = v_1 + v_2 \quad \text{(Equation 1)} \]

### Step 4: Apply Conservation of Energy
The total mechanical energy before and after the collision must also be conserved. The initial kinetic energy is:
\[ \text{Initial KE} = \frac{1}{2} m u^2 \]
The final kinetic energy plus the energy used for ionization is:
\[ \text{Final KE} = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 + \Delta E \]
Setting these equal gives:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 + \Delta E \quad \text{(Equation 2)} \]

### Step 5: Substitute and Rearrange
From Equation 1, we can express \( v_2 \) in terms of \( u \) and \( v_1 \):
\[ v_2 = u - v_1 \]
Substituting \( v_2 \) into Equation 2:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m (u - v_1)^2 + \Delta E \]
Expanding the equation:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m (u^2 - 2uv_1 + v_1^2) + \Delta E \]
This simplifies to:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m u^2 - m uv_1 + \Delta E \]
Cancelling \( \frac{1}{2} m u^2 \) from both sides gives:
\[ 0 = -m uv_1 + \Delta E \]
Rearranging gives:
\[ m uv_1 = \Delta E \]
Thus:
\[ v_1 = \frac{\Delta E}{m u} \]

### Step 6: Minimum Condition
To find the minimum value of \( u \), we can set \( v_1 = v_2 \) (for maximum energy transfer):
\[ u = 2v_1 \]
Substituting \( v_1 \):
\[ u = 2 \cdot \frac{\Delta E}{m u} \]
Squaring both sides:
\[ u^2 = \frac{4 \Delta E}{m} \]
Thus:
\[ u = \sqrt{\frac{4 \Delta E}{m}} \]

### Step 7: Substitute Values
Substituting the values of \( \Delta E \) and \( m \):
\[ u = \sqrt{\frac{4 \times 2.176 \times 10^{-18}}{1.67 \times 10^{-27}}} \]
Calculating gives:
\[ u = \sqrt{\frac{8.704 \times 10^{-18}}{1.67 \times 10^{-27}}} \approx 7.2 \times 10^4 \, \text{m/s} \]

### Final Answer
The minimum speed \( u \) required for one of the hydrogen atoms to get ionized is:
\[ \boxed{7.2 \times 10^4 \, \text{m/s}} \]