The circumference of first orbit of hydrogen atom is .S.. Then the Broglie wavelength of electron in that orbit is

Options

A. `S//2`
B. 2S
C. S
D. 3S

Answer

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

### Step 1: Understand the relationship between circumference and radius
The circumference \( C \) of the first orbit is given as \( S \). The formula for the circumference of a circle is:
\[ C = 2\pi R \]
For the first orbit, we can express this as:
\[ S = 2\pi R_1 \]
where \( R_1 \) is the radius of the first orbit.

### Step 2: Relate angular momentum to the de Broglie wavelength
According to Bohr's model, the angular momentum \( L \) of the electron in the nth orbit is quantized and given by:
\[ L_n = n\frac{h}{2\pi} \]
For the first orbit (where \( n = 1 \)):
\[ L_1 = \frac{h}{2\pi} \]

### Step 3: Express angular momentum in terms of linear momentum
The angular momentum can also be expressed as:
\[ L = m v R \]
For the first orbit, this gives:
\[ L_1 = m v_1 R_1 \]
Setting the two expressions for angular momentum equal, we have:
\[ m v_1 R_1 = \frac{h}{2\pi} \]

### Step 4: Solve for \( v_1 R_1 \)
Rearranging the equation gives:
\[ v_1 R_1 = \frac{h}{2\pi m} \]

### Step 5: Find the de Broglie wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting \( v_1 \) into this equation, we can express the de Broglie wavelength in terms of the radius:
\[ \lambda = \frac{h}{mv_1} \]

### Step 6: Relate \( v_1 \) to the circumference
From the earlier steps, we know that:
\[ 2\pi R_1 = S \]
Thus, we can express the de Broglie wavelength as:
\[ \lambda = 2\pi R_1 \]
Since \( 2\pi R_1 = S \), we have:
\[ \lambda = S \]

### Conclusion
Thus, the de Broglie wavelength of the electron in the first orbit of the hydrogen atom is:
\[ \lambda = S \]