The de-Broglie's wavelength of an electron in first orbit of Bohr's hydrogen is equal to

Options

A. Radius of the orbit
B. Perimeter of the orbit
C. Diameter of the orbit
D. Half of the perimeter of the orbit

Explanation

To find the de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen model, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

### Step 2: Express momentum in terms of mass and velocity
The momentum \( p \) of an electron can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{mv} \]

### Step 4: Use Bohr's model to relate angular momentum and velocity
According to Bohr's model, the angular momentum \( L \) of an electron in the nth orbit is given by:
\[ L = mvr = \frac{nh}{2\pi} \]
For the first orbit (n=1), this becomes:
\[ mvr = \frac{h}{2\pi} \]

### Step 5: Solve for velocity
From the angular momentum equation, we can express \( mv \) as:
\[ mv = \frac{h}{2\pi r} \]

### Step 6: Substitute back into the de-Broglie wavelength formula
Now substitute \( mv \) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\frac{h}{2\pi r}} = 2\pi r \]

### Step 7: Interpret the result
The expression \( 2\pi r \) represents the circumference (perimeter) of the orbit of the electron in the first orbit of hydrogen.

### Final Answer
Thus, the de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen is equal to the perimeter of the orbit.