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 `He^(+)` ion

Answer

To find the de Broglie wavelength of an electron in the second orbit of the \( \text{Li}^{2+} \) ion and compare it with other options, 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 electron,
- \( v \) is the velocity of the electron.

### Step 2: Calculate the Velocity of the Electron in the Second Orbit of \( \text{Li}^{2+} \)
For a hydrogen-like atom, the velocity of the electron in the \( n \)-th orbit can be expressed as:
\[ v = \frac{2.16 \times 10^6 \cdot Z}{n} \]
where:
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

For \( \text{Li}^{2+} \):
- \( Z = 3 \) (since lithium has an atomic number of 3),
- \( n = 2 \).

Substituting these values:
\[ v = \frac{2.16 \times 10^6 \cdot 3}{2} = 3.24 \times 10^6 \, \text{m/s} \]

### Step 3: Substitute into the de Broglie Wavelength Formula
Now we can substitute the velocity back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{m \cdot v} \]
Using the previously calculated velocity:
\[ \lambda = \frac{h}{m \cdot 3.24 \times 10^6} \]

### Step 4: Compare with Other Options
To find which other electron's de Broglie wavelength is equal to this, we need to check the de Broglie wavelength for electrons in other hydrogen-like ions.

#### For \( n = 3 \) of a hydrogen-like atom:
Using the formula for velocity:
\[ v = \frac{2.16 \times 10^6 \cdot Z}{3} \]
Setting \( Z = 1 \) (for hydrogen):
\[ v = \frac{2.16 \times 10^6 \cdot 1}{3} = 7.2 \times 10^5 \, \text{m/s} \]
Calculating \( \lambda \):
\[ \lambda = \frac{h}{m \cdot 7.2 \times 10^5} \]
This will not equal the previous \( \lambda \).

#### For \( n = 4 \) of \( \text{C}^{5+} \):
Using the formula for velocity:
\[ v = \frac{2.16 \times 10^6 \cdot Z}{4} \]
Setting \( Z = 6 \) (for carbon):
\[ v = \frac{2.16 \times 10^6 \cdot 6}{4} = 3.24 \times 10^6 \, \text{m/s} \]
Calculating \( \lambda \):
\[ \lambda = \frac{h}{m \cdot 3.24 \times 10^6} \]
This matches the \( \lambda \) we calculated for \( \text{Li}^{2+} \).

### Conclusion
The de Broglie wavelength of the electron in the second orbit of \( \text{Li}^{2+} \) is equal to the de Broglie wavelength of the electron in the fourth orbit of \( \text{C}^{5+} \).