In the nth orbit of hydrogen atom, find the ratio of the radius of the electron orbit and de-Broglie wavelength associated with it.

Options

A. `(n)/(2pi)`
B. `(n^(2))/(2pi)`
C. `(1)/(2pin)`
D. `(1)/(2pin^(2))`

Answer

To find the ratio of the radius of the electron orbit to the de-Broglie wavelength associated with it in the nth orbit of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (λ) of an electron 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. **Angular Momentum in Bohr's Model**:
According to Bohr's model, the angular momentum (L) of the electron in the nth orbit is quantized and given by:
\[ L = mvr = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number, \( r \) is the radius of the orbit, and \( v \) is the velocity of the electron.

3. **Expressing Momentum**:
From the angular momentum equation, we can express \( mv \) as:
\[ mv = \frac{n h}{2\pi r} \]

4. **Substituting into de-Broglie Wavelength**:
Now, substitute \( mv \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{n h}{2\pi r}} = \frac{2\pi r}{n} \]

5. **Finding the Ratio**:
We need to find the ratio of the radius \( r \) to the wavelength \( \lambda \):
\[ \frac{r}{\lambda} = \frac{r}{\frac{2\pi r}{n}} = \frac{n}{2\pi} \]

6. **Final Result**:
Therefore, the ratio of the radius of the electron orbit to the de-Broglie wavelength associated with it in the nth orbit of a hydrogen atom is:
\[ \frac{r}{\lambda} = \frac{n}{2\pi} \]

### Conclusion:
The final answer is:
\[ \frac{r}{\lambda} = \frac{n}{2\pi} \]