The de-Broglie wavelength `(lambda_(B))` associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state `(lambda_(G))` by :

Options

A. `lambda_(B)=2lambda_(G)`
B. `lambda_(B)=lambda_(G//3)`
C. `lambda_(B)=3lambda_(G)`
D. `lambda_(B)=lambda_(G//2)`

Answer

To solve the problem of finding the relationship between the de-Broglie wavelengths associated with the electron in the second excited state and the ground state of a hydrogen atom, we will follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (λ) 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 angular momentum to quantum states
For an electron in a hydrogen atom, the angular momentum (L) is quantized and given by:
\[ L = mvr = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number.

### Step 3: Express the de-Broglie wavelength in terms of quantum number
From the angular momentum equation, we can express the velocity \( v \) as:
\[ v = \frac{n h}{2\pi m r} \]
Substituting this expression for \( v \) into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m \left(\frac{n h}{2\pi m r}\right)} = \frac{2\pi r}{n} \]

### Step 4: Determine the radius of the electron's orbit
For a hydrogen-like atom, the radius \( r \) of the electron's orbit is given by:
\[ r = a_0 \frac{n^2}{z} \]
where \( a_0 \) is the Bohr radius and \( z \) is the atomic number (which is 1 for hydrogen).

### Step 5: Substitute the radius into the wavelength formula
Substituting the expression for \( r \) into the wavelength formula:
\[ \lambda = \frac{2\pi \left(a_0 \frac{n^2}{z}\right)}{n} = \frac{2\pi a_0 n}{z} \]
For hydrogen, \( z = 1 \), thus:
\[ \lambda = 2\pi a_0 n \]

### Step 6: Calculate the wavelengths for the second excited state and ground state
- For the ground state (\( n = 1 \)):
\[ \lambda_G = 2\pi a_0 (1) = 2\pi a_0 \]

- For the second excited state (\( n = 3 \)):
\[ \lambda_{3} = 2\pi a_0 (3) = 6\pi a_0 \]

### Step 7: Find the relationship between the two wavelengths
Now, we can find the relationship between the de-Broglie wavelength in the second excited state and the ground state:
\[ \frac{\lambda_{3}}{\lambda_G} = \frac{6\pi a_0}{2\pi a_0} = 3 \]
Thus, the de-Broglie wavelength associated with the electron in the second excited state is three times that in the ground state.

### Final Answer
\[ \lambda_{3} = 3 \lambda_{G} \]