A hydrogen atom in ground state, moving with speed `v` collides with another hydrogen atom in ground state at rest. If `vlev_(0)=sqrt((kE_(0))/(m_(H)))~~a*bcxx10^(n)m//s`, then the collision is elastic. Here `a,b` and `c` are whole number, less then 9. Find the value of `(axxbxxc)/(kxxn)`
`(me^(4))/(2h^(2))=E_(0)=13.6eV//"atom"=2.18xx10^(-18)J//"atom" implies` ionisation energy of `H` -atom
`implies m_(H)=1.67xx10^(-27)kgimplies` Mass of hydrogen atom.

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the necessary values.

### Step 1: Understand the Problem
We have a hydrogen atom moving with speed \( v \) colliding with another hydrogen atom at rest. The condition for the collision to be elastic is given as:
\[ v \leq v_0 = \sqrt{\frac{k E_0}{m_H}} \]
where \( E_0 \) is the ionization energy of the hydrogen atom, \( m_H \) is the mass of the hydrogen atom, and \( k \) is a constant.

### Step 2: Given Constants
We know:
- Ionization energy of hydrogen atom:
\[ E_0 = 13.6 \, \text{eV} = 2.18 \times 10^{-18} \, \text{J} \]
- Mass of hydrogen atom:
\[ m_H = 1.67 \times 10^{-27} \, \text{kg} \]

### Step 3: Calculate \( v_0 \)
Using the formula for \( v_0 \):
\[ v_0 = \sqrt{\frac{k E_0}{m_H}} \]
We need to express \( v_0 \) in the form \( a.bc \times 10^n \, \text{m/s} \).

### Step 4: Substitute Values
Substituting the known values into the equation:
\[ v_0 = \sqrt{\frac{k \times 2.18 \times 10^{-18}}{1.67 \times 10^{-27}}} \]
This simplifies to:
\[ v_0 = \sqrt{\frac{2.18 \times 10^{-18} k}{1.67 \times 10^{-27}}} \]

### Step 5: Simplify the Expression
Calculating the fraction:
\[ \frac{2.18}{1.67} \approx 1.303 \]
Thus:
\[ v_0 \approx \sqrt{1.303 \times 10^9 \cdot k} \]

### Step 6: Express in Scientific Notation
Assuming \( k = 3 \) (as inferred from the video), we have:
\[ v_0 \approx \sqrt{1.303 \times 10^9 \times 3} = \sqrt{3.909 \times 10^9} \approx 6.26 \times 10^4 \, \text{m/s} \]

### Step 7: Identify \( a, b, c, n \)
From the expression \( 6.26 \times 10^4 \):
- \( a = 6 \)
- \( b = 2 \)
- \( c = 6 \)
- \( n = 4 \)

### Step 8: Calculate \( \frac{a \times b \times c}{k \times n} \)
Now, substituting the values:
\[ \frac{a \times b \times c}{k \times n} = \frac{6 \times 2 \times 6}{3 \times 4} \]
Calculating the numerator:
\[ 6 \times 2 \times 6 = 72 \]
Calculating the denominator:
\[ 3 \times 4 = 12 \]
Thus:
\[ \frac{72}{12} = 6 \]

### Final Answer
The value of \( \frac{a \times b \times c}{k \times n} \) is \( 6 \).