If the radius of firs Bohr's orbit is x, then de-Broglie wavelenght of electron in 3rd orbit is nearly `(npix)` Find value of n

Answer

To solve the problem, we need to find the value of \( n \) such that the de-Broglie wavelength of an electron in the 3rd Bohr orbit is approximately \( n \pi x \), where \( x \) is the radius of the first Bohr orbit.

### Step-by-Step Solution:

1. **Understanding the Radius of Bohr Orbits**:
The radius of the \( n \)-th Bohr orbit is given by the formula:
\[ r_n = n^2 r_1 \]
where \( r_1 \) is the radius of the first orbit. Given that the radius of the first orbit is \( x \), we have:
\[ r_1 = x \quad \text{and} \quad r_3 = 3^2 r_1 = 9x \]

2. **Using the de-Broglie Wavelength Formula**:
The de-Broglie wavelength \( \lambda \) of an electron in the \( n \)-th orbit is given by:
\[ \lambda = \frac{2 \pi r_n}{n} \]
Substituting \( r_3 \) into the equation, we get:
\[ \lambda_3 = \frac{2 \pi (9x)}{3} \]

3. **Calculating the Wavelength**:
Simplifying the expression for \( \lambda_3 \):
\[ \lambda_3 = \frac{18 \pi x}{3} = 6 \pi x \]

4. **Setting Up the Equation**:
We need to find \( n \) such that:
\[ \lambda_3 \approx n \pi x \]
From our calculation, we have:
\[ 6 \pi x = n \pi x \]

5. **Solving for \( n \)**:
Dividing both sides by \( \pi x \) (assuming \( \pi x \neq 0 \)):
\[ n = 6 \]

### Final Answer:
Thus, the value of \( n \) is \( 6 \).

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