de-Broglie wavelength of electron in `2^("nd")` excited state of hydrogen atom is: [where `r_(0)` is the radius of `1^("st")` orbit in H-atom]

Options

A. `r_(0)`
B. `pir_(0)`
C. `3pir_(0)`
D. `6pir_(0)`

Explanation

To find the de-Broglie wavelength of an electron in the \(2^{nd}\) excited state 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:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(v\) is its velocity.

### Step 2: Relate momentum to the quantum state
The momentum (\(p\)) of an electron in a hydrogen atom can also be expressed in terms of quantum numbers:
\[ p = mv = \frac{n h}{2 \pi r} \]
where \(n\) is the principal quantum number and \(r\) is the radius of the orbit.

### Step 3: Substitute momentum into the de-Broglie wavelength formula
Substituting the expression for momentum into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{p} = \frac{h}{\frac{n h}{2 \pi r}} = \frac{2 \pi r}{n} \]

### Step 4: Determine the radius for the \(2^{nd}\) excited state
For the hydrogen atom, the radius of the \(n^{th}\) orbit is given by:
\[ r_n = n^2 r_0 \]
where \(r_0\) is the radius of the first orbit. For the \(2^{nd}\) excited state, \(n = 3\) (since the ground state is \(n=1\), the first excited state is \(n=2\), and the second excited state is \(n=3\)):
\[ r_3 = 3^2 r_0 = 9 r_0 \]

### Step 5: Substitute \(n\) and \(r\) into the wavelength formula
Now substituting \(n = 3\) and \(r = 9 r_0\) into the wavelength formula:
\[ \lambda = \frac{2 \pi (9 r_0)}{3} = 6 \pi r_0 \]

### Final Answer
Thus, the de-Broglie wavelength of the electron in the \(2^{nd}\) excited state of the hydrogen atom is:
\[ \lambda = 6 \pi r_0 \]