The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

Options

A. `(0.529)/(2pi)Å`
B. `2pixx0.529Å`
C. `0.529Å`
D. `4xx0.529Å`

Explanation

To find the de Broglie's wavelength of the electron present in the first Bohr orbit of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for de Broglie's Wavelength**:
The de Broglie wavelength (λ) is related to the radius (r) of the orbit and the principal quantum number (n) by the formula:
\[ 2\pi r = n\lambda \]
Rearranging this gives:
\[ \lambda = \frac{2\pi r}{n} \]

2. **Determine the Radius of the First Bohr Orbit**:
The radius of the nth Bohr orbit is given by the formula:
\[ r_n = 0.529 \times n^2 \times \frac{1}{Z} \text{ angstroms} \]
For hydrogen (Z = 1) and the first orbit (n = 1):
\[ r_1 = 0.529 \times 1^2 \times \frac{1}{1} = 0.529 \text{ angstroms} \]

3. **Substitute Values into the Wavelength Formula**:
Now, substituting the values into the de Broglie wavelength formula:
\[ \lambda = \frac{2\pi \times 0.529 \text{ angstroms}}{1} \]
This simplifies to:
\[ \lambda = 2\pi \times 0.529 \text{ angstroms} \]

4. **Calculate the Wavelength**:
Using the approximate value of π (3.14):
\[ \lambda \approx 2 \times 3.14 \times 0.529 \approx 3.34 \text{ angstroms} \]

5. **Final Answer**:
The de Broglie's wavelength of the electron in the first Bohr orbit of the hydrogen atom is approximately:
\[ \lambda \approx 3.34 \text{ angstroms} \]