The de-Broglie wavelength `lambda_(n)`of the electron in the `n^(th)` orbit of hydrogen atom is

Options

A. inversely to n
B. Proportional to `n^(2)`
C. Proportional to n
D. Inversely proportional to `n^(2)`

Answer

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

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

### Step 2: Express momentum in terms of mass and velocity
The momentum \(p\) of the 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, we get:
\[ \lambda_n = \frac{h}{mv} \]

### Step 4: Use Bohr's model to relate velocity and radius
According to Bohr's model of the hydrogen atom, the angular momentum \(L\) of the electron in the \(n^{th}\) orbit is quantized and given by:
\[ L = mvr = n \frac{h}{2\pi} \]
where \(r\) is the radius of the \(n^{th}\) orbit and \(n\) is the principal quantum number.

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

### Step 6: Substitute \(mv\) back into the de-Broglie wavelength formula
Now, substituting this expression for \(mv\) back into the de-Broglie wavelength formula gives:
\[ \lambda_n = \frac{h}{n \frac{h}{2\pi r}} = \frac{2\pi r}{n} \]

### Step 7: Analyze the relationship
From the equation \(\lambda_n = \frac{2\pi r}{n}\), we can see that the de-Broglie wavelength \(\lambda_n\) is inversely proportional to the principal quantum number \(n\).

### Conclusion
Thus, we conclude that the de-Broglie wavelength of the electron in the \(n^{th}\) orbit of hydrogen is given by:
\[ \lambda_n \propto \frac{1}{n} \]