The circumference of `n^(th)` orbit in H-atom can be expressed in terms of deBroglie wavelength ` lambda` as :

Options

A. `(0.529)nlambda`
B. `sqrt(nlambda)`
C. `(13.6)lambda`
D. `nlambda`

Explanation

To find the circumference of the \( n^{th} \) orbit in a hydrogen atom in terms of the de Broglie wavelength \( \lambda \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is its velocity.

2. **Circumference of the Orbit**:
According to the Bohr model, the circumference \( C \) of the \( n^{th} \) orbit can be expressed as:
\[ C = 2\pi r \]
where \( r \) is the radius of the \( n^{th} \) orbit.

3. **Relating Circumference to Angular Momentum**:
The angular momentum of the electron in the \( n^{th} \) orbit is quantized and given by:
\[ mvr = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number.

4. **Substituting for \( mv \)**:
From the angular momentum equation, we can express \( mv \) as:
\[ mv = \frac{nh}{2\pi r} \]

5. **Substituting into the Circumference Equation**:
Now, substituting \( mv \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \]

6. **Expressing Circumference in Terms of \( \lambda \)**:
Rearranging the equation gives us:
\[ 2\pi r = n\lambda \]
Thus, the circumference of the \( n^{th} \) orbit can be expressed as:
\[ C = 2\pi r = n\lambda \]

### Final Answer:
The circumference of the \( n^{th} \) orbit in a hydrogen atom can be expressed as:
\[ C = n\lambda \]