de Broglie wavelength of electron in second orbit of `Li^(2+)` ion will be equal to de Broglie's wavelength of electron in :

Options

A. `n=3` of `H-` atom
B. `n=4` of` C^(5+)` ion
C. `n=6` of `Be^(3+)` ion
D. `n=3` of `H^(+)` ion

Explanation

To determine the de Broglie wavelength of an electron in the second orbit of a \( \text{Li}^{2+} \) ion and compare it with the de Broglie wavelength of an electron in another ion, we need to use the relationship between the principal quantum number \( n \) and the atomic number \( Z \).

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength:**
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is the velocity of the electron.

2. **Bohr's Postulate:**
According to Bohr's postulate, the angular momentum \( L \) of an electron in an orbit is quantized and given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where \( n \) is the principal quantum number.

3. **Radius of the nth Orbit:**
The radius \( r_n \) of the nth orbit in a hydrogen-like atom is given by:
\[ r_n = \frac{0.53 \, \text{Å} \cdot n^2}{Z} \]
where \( Z \) is the atomic number.

4. **De Broglie Wavelength and Radius Relationship:**
Using the above relationships, we can express the de Broglie wavelength \( \lambda \) in terms of \( n \) and \( Z \):
\[ \lambda \propto \frac{n}{Z} \]

5. **Given Conditions:**
For \( \text{Li}^{2+} \) ion:
- Atomic number \( Z = 3 \)
- Principal quantum number \( n = 2 \)

Therefore, the ratio \( \frac{n}{Z} \) for \( \text{Li}^{2+} \) is:
\[ \frac{n}{Z} = \frac{2}{3} \]

6. **Comparing with Other Ions:**
We need to find another ion where the ratio \( \frac{n}{Z} \) is the same.

- For \( \text{Be}^{3+} \) (Beryllium ion with atomic number 4):
\[ \frac{n}{Z} = \frac{6}{4} = \frac{3}{2} \]
This does not match \( \frac{2}{3} \).

- For \( \text{C}^{5+} \) (Carbon ion with atomic number 6):
\[ \frac{n}{Z} = \frac{2}{6} = \frac{1}{3} \]
This does not match \( \frac{2}{3} \).

- For \( \text{B}^{4+} \) (Boron ion with atomic number 5):
\[ \frac{n}{Z} = \frac{3}{5} \]
This does not match \( \frac{2}{3} \).

- For \( \text{C}^{5+} \) (Carbon ion with atomic number 6):
\[ \frac{n}{Z} = \frac{2}{3} \]
This matches \( \frac{2}{3} \).

Therefore, the de Broglie wavelength of the electron in the second orbit of \( \text{Li}^{2+} \) ion is equal to the de Broglie wavelength of the electron in the second orbit of \( \text{C}^{5+} \) ion.