Answer :
To determine the de-Broglie wavelength of the electron in the second orbit of a [tex]\( \text{Li}^{2+} \)[/tex] ion, we'll use the specific given conditions and known fundamental constants.
### Step-by-Step Solution:
1. Determine the relevant constants:
- Bohr radius [tex]\( a_0 \)[/tex]: This is given as [tex]\( 52.9 \, \text{pm} \)[/tex]. This is the average distance of the electron from the nucleus in the hydrogen atom's ground state.
- Atomic number [tex]\( Z \)[/tex] of [tex]\( \text{Li}^{2+} \)[/tex]: For lithium, [tex]\( Z = 3 \)[/tex]. For the [tex]\( \text{Li}^{2+} \)[/tex] ion, this is the atomic number since it has lost two electrons.
- Orbit number [tex]\( n \)[/tex]: We are considering the second orbit, meaning [tex]\( n = 2 \)[/tex].
2. Formula for the de-Broglie wavelength [tex]\( \lambda \)[/tex]:
The de-Broglie wavelength is expressed as:
[tex]\[ \lambda = \frac{2 \pi n a_0}{Z} \][/tex]
Here, [tex]\( \lambda \)[/tex] is the de-Broglie wavelength, [tex]\( n \)[/tex] is the orbit number, [tex]\( a_0 \)[/tex] is the Bohr radius, and [tex]\( Z \)[/tex] is the atomic number.
3. Substituting the given values into the formula:
[tex]\[ \lambda = \frac{2 \pi \cdot 2 \cdot 52.9 \, \text{pm}}{3} \][/tex]
4. Calculating the wavelength [tex]\( \lambda \)[/tex]:
Plug in the values:
[tex]\[ \lambda = \frac{2 \cdot \pi \cdot 2 \cdot 52.9}{3} \][/tex]
5. Perform the multiplication and division:
- First, multiply [tex]\( 2 \cdot 52.9 = 105.8 \)[/tex].
- Then, multiply by [tex]\( 2 \pi \)[/tex]:
[tex]\[ 105.8 \cdot 2 \pi \approx 105.8 \cdot 6.2832 \approx 664.51336 \][/tex]
- Finally, divide by [tex]\( 3 \)[/tex]:
[tex]\[ \lambda \approx \frac{664.51336}{3} \approx 221.50445 \][/tex]
### Conclusion:
Hence, the de-Broglie wavelength of the electron in the second orbit of the [tex]\( \text{Li}^{2+} \)[/tex] ion is approximately [tex]\( 221.587 \, \text{pm} \)[/tex].
### Step-by-Step Solution:
1. Determine the relevant constants:
- Bohr radius [tex]\( a_0 \)[/tex]: This is given as [tex]\( 52.9 \, \text{pm} \)[/tex]. This is the average distance of the electron from the nucleus in the hydrogen atom's ground state.
- Atomic number [tex]\( Z \)[/tex] of [tex]\( \text{Li}^{2+} \)[/tex]: For lithium, [tex]\( Z = 3 \)[/tex]. For the [tex]\( \text{Li}^{2+} \)[/tex] ion, this is the atomic number since it has lost two electrons.
- Orbit number [tex]\( n \)[/tex]: We are considering the second orbit, meaning [tex]\( n = 2 \)[/tex].
2. Formula for the de-Broglie wavelength [tex]\( \lambda \)[/tex]:
The de-Broglie wavelength is expressed as:
[tex]\[ \lambda = \frac{2 \pi n a_0}{Z} \][/tex]
Here, [tex]\( \lambda \)[/tex] is the de-Broglie wavelength, [tex]\( n \)[/tex] is the orbit number, [tex]\( a_0 \)[/tex] is the Bohr radius, and [tex]\( Z \)[/tex] is the atomic number.
3. Substituting the given values into the formula:
[tex]\[ \lambda = \frac{2 \pi \cdot 2 \cdot 52.9 \, \text{pm}}{3} \][/tex]
4. Calculating the wavelength [tex]\( \lambda \)[/tex]:
Plug in the values:
[tex]\[ \lambda = \frac{2 \cdot \pi \cdot 2 \cdot 52.9}{3} \][/tex]
5. Perform the multiplication and division:
- First, multiply [tex]\( 2 \cdot 52.9 = 105.8 \)[/tex].
- Then, multiply by [tex]\( 2 \pi \)[/tex]:
[tex]\[ 105.8 \cdot 2 \pi \approx 105.8 \cdot 6.2832 \approx 664.51336 \][/tex]
- Finally, divide by [tex]\( 3 \)[/tex]:
[tex]\[ \lambda \approx \frac{664.51336}{3} \approx 221.50445 \][/tex]
### Conclusion:
Hence, the de-Broglie wavelength of the electron in the second orbit of the [tex]\( \text{Li}^{2+} \)[/tex] ion is approximately [tex]\( 221.587 \, \text{pm} \)[/tex].
