To solve the problem of finding the energy transition for [tex]\( Be^{3+} \)[/tex] from [tex]\( n=1 \)[/tex] to [tex]\( n=2 \)[/tex] given that the energy transition for Hydrogen (H) in the same states is 10.2 eV, we follow these steps:
1. Understand the formula for energy transition:
The energy transition for an electron moving between two energy levels in a hydrogen-like atom can be calculated using the formula:
[tex]\[
E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
where:
- [tex]\( E \)[/tex] is the energy transition
- [tex]\( Z \)[/tex] is the atomic number of the atom
- [tex]\( n_1 \)[/tex] is the initial energy level
- [tex]\( n_2 \)[/tex] is the final energy level
2. Identify the given values:
- For Hydrogen (H), the atomic number [tex]\( Z_H = 1 \)[/tex].
- For [tex]\( Be^{3+} \)[/tex], the atomic number [tex]\( Z_{Be} = 4 \)[/tex] because Beryllium (Be) has an atomic number of 4.
3. Energy transition in Hydrogen (H):
We know from the problem that the transition energy for Hydrogen from [tex]\( n=1 \)[/tex] to [tex]\( n=2 \)[/tex] is 10.2 eV.
4. Determine the transition ratio:
The energy levels in hydrogen-like atoms scale with [tex]\( Z^2 \)[/tex]. Therefore, the ratio of the energy transitions between [tex]\( Be^{3+} \)[/tex] and H can be calculated as:
[tex]\[
\text{transition ratio} = \left( \frac{Z_{Be}}{Z_H} \right)^2 = \left( \frac{4}{1} \right)^2 = 16
\][/tex]
5. Calculate the energy transition for [tex]\( Be^{3+} \)[/tex]:
Using the transition ratio, we multiply the given energy transition for Hydrogen by this ratio to find the energy for [tex]\( Be^{3+} \)[/tex]:
[tex]\[
E_{Be^{3+}} = 10.2 \, \text{eV} \times 16 = 163.2 \, \text{eV}
\][/tex]
6. Conclusion:
The energy for the transition from [tex]\( n=1 \)[/tex] to [tex]\( n=2 \)[/tex] in [tex]\( Be^{3+} \)[/tex] is therefore [tex]\( 163.2 \)[/tex] eV.
Thus, the correct answer is:
[tex]\[ \boxed{163.2 \text{ eV}} \][/tex]
So, the correct option is (2) 163.2 eV.
