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 \(\lambda\) 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 the electron can be expressed as:
\[ p = mv \]
where \(m\) is the mass of the electron and \(v\) is its velocity.

### Step 3: Determine the velocity of the electron in the \(n^{th}\) orbit
For an electron in the \(n^{th}\) orbit of a hydrogen atom, the velocity \(v_n\) can be derived from the Bohr model. The velocity is given by:
\[ v_n = \frac{2 \pi^2 m z}{n h} \cdot \frac{1}{4 \pi \epsilon_0} \]
where \(z\) is the atomic number (which is 1 for hydrogen), \(n\) is the principal quantum number, and \(\epsilon_0\) is the permittivity of free space.

### Step 4: Analyze the relationship between wavelength and orbit number
From the above expression, we can see that the velocity \(v_n\) is inversely proportional to \(n\):
\[ v_n \propto \frac{1}{n} \]
Since momentum \(p\) is directly proportional to velocity:
\[ p \propto v_n \propto \frac{1}{n} \]
This implies that:
\[ \lambda \propto \frac{1}{p} \propto n \]

### Step 5: Conclusion
Thus, we conclude that the de-Broglie wavelength \(\lambda_n\) of the electron in the \(n^{th}\) orbit of the hydrogen atom is directly proportional to \(n\):
\[ \lambda_n \propto n \]

### Final Answer
The de-Broglie wavelength \(\lambda_n\) of the electron in the \(n^{th}\) orbit of the hydrogen atom is proportional to \(n\).

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