A H-atom moving with speed v makes a head on collisioon with a H-atom at rest . Both atoms are in ground state. The minimum value of velocity v for which one of the atom may excite is `(1.25 nxx10^(4))` m/s Find value of x.

Answer

To solve the problem, we need to find the minimum speed \( v \) of a hydrogen atom required to excite another hydrogen atom at rest during a head-on collision. Here’s a step-by-step solution:

### Step 1: Understand the Energy Requirement for Excitation
The minimum energy required to excite a hydrogen atom from the ground state (n=1) to the first excited state (n=2) is given by the energy difference between these two states. This energy difference is approximately \( 10.2 \, \text{eV} \).

### Step 2: Convert Energy from eV to Joules
To use this energy in our calculations, we need to convert it from electron volts (eV) to joules (J). The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus, the energy in joules is:
\[ \Delta E = 10.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.632 \times 10^{-18} \, \text{J} \]

### Step 3: Apply Conservation of Momentum
In a perfectly elastic collision, the momentum before and after the collision is conserved. Let \( m \) be the mass of the hydrogen atom. Before the collision, the momentum is:
\[ p_{\text{initial}} = mv \]
After the collision, both atoms will have the same velocity \( V \):
\[ p_{\text{final}} = 2mV \]
Setting the initial and final momentum equal gives:
\[ mv = 2mV \implies V = \frac{v}{2} \]

### Step 4: Apply Conservation of Energy
The kinetic energy before the collision must equal the kinetic energy after the collision plus the energy required to excite the atom:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mV^2 + \Delta E \]
Substituting \( V = \frac{v}{2} \) into the equation:
\[ \frac{1}{2} mv^2 = \frac{1}{2} m\left(\frac{v}{2}\right)^2 + \Delta E \]
This simplifies to:
\[ \frac{1}{2} mv^2 = \frac{1}{8} mv^2 + \Delta E \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{1}{2} mv^2 - \frac{1}{8} mv^2 = \Delta E \]
Combining the terms on the left:
\[ \left(\frac{4}{8} - \frac{1}{8}\right) mv^2 = \Delta E \]
This simplifies to:
\[ \frac{3}{8} mv^2 = \Delta E \]

### Step 6: Solve for \( v \)
Now, we can solve for \( v^2 \):
\[ v^2 = \frac{8\Delta E}{3m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{8\Delta E}{3m}} \]

### Step 7: Substitute Values
Substituting \( \Delta E = 1.632 \times 10^{-18} \, \text{J} \) and the mass of the hydrogen atom \( m = 1.67 \times 10^{-27} \, \text{kg} \):
\[ v = \sqrt{\frac{8 \times 1.632 \times 10^{-18}}{3 \times 1.67 \times 10^{-27}}} \]

### Step 8: Calculate \( v \)
Calculating the above expression:
\[ v = \sqrt{\frac{1.3056 \times 10^{-17}}{5.01 \times 10^{-27}}} \approx \sqrt{2.605 \times 10^{9}} \approx 5.1 \times 10^{4} \, \text{m/s} \]

### Step 9: Compare with Given Value
The problem states that the minimum velocity \( v \) is \( 1.25 \times x \times 10^{4} \, \text{m/s} \). Setting this equal to our calculated value:
\[ 1.25x \times 10^{4} = 5.1 \times 10^{4} \]
Dividing both sides by \( 1.25 \times 10^{4} \):
\[ x = \frac{5.1}{1.25} = 4.08 \]

### Final Answer
Thus, the value of \( x \) is approximately \( 4 \).