Consider Bohr's theory for hydrogen atom . The magnitude of orbit angular momentum orbit radius and velocity of the electron in nth energy state in a hydrogen atom are l , r and v respectively. Find out the value of 'x' if product of v, r and l (vrl) is directly proportional to `n^(x)`.

Explanation

To solve the problem, we will use Bohr's theory for the hydrogen atom to find the relationship between the angular momentum (L), orbit radius (R), and velocity (V) of the electron in the nth energy state. We will then determine the value of 'x' in the expression \( vrl \propto n^x \).

### Step-by-Step Solution:

1. **Angular Momentum (L)**:
According to Bohr's theory, the angular momentum of an electron in the nth orbit is given by:
\[ L = \frac{n h}{2 \pi} \]
where \( n \) is the principal quantum number and \( h \) is Planck's constant.

2. **Orbit Radius (R)**:
The radius of the nth orbit is given by:
\[ R = n^2 R_0 \]
where \( R_0 \) is the radius of the ground state (n=1).

3. **Velocity (V)**:
The velocity of the electron in the nth orbit is given by:
\[ V = \frac{2.18 \times 10^6 \, \text{m/s} \cdot Z}{n} \]
where \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)).

4. **Product of v, r, and l**:
Now we need to find the product \( vrl \):
\[ vrl = V \cdot R \cdot L \]
Substituting the expressions we derived:
\[ vrl = \left(\frac{2.18 \times 10^6 \cdot Z}{n}\right) \cdot (n^2 R_0) \cdot \left(\frac{n h}{2 \pi}\right) \]

5. **Simplifying the Expression**:
Now, substituting and simplifying:
\[ vrl = \left(\frac{2.18 \times 10^6 \cdot Z \cdot n^2 R_0 \cdot n h}{2 \pi n}\right) \]
\[ vrl = \frac{2.18 \times 10^6 \cdot Z \cdot R_0 \cdot h}{2 \pi} \cdot n^2 \]
This shows that \( vrl \) is directly proportional to \( n^2 \).

6. **Finding x**:
From the above expression, we can conclude that:
\[ vrl \propto n^2 \]
Therefore, the value of \( x \) is:
\[ x = 2 \]

### Final Answer:
The value of \( x \) is \( 2 \).