Question

Among the given options choose the correct energy of transition:

• A. $H_{2\to1}$ $(6.8\text{ eV})$
• B. $Li^{2+}_{2\to1}$ $(13.6\text{ eV})$
• C. $He^{+}_{2\to1}$ $(40.8\text{ eV})$
• D. $Be^{3+}_{2\to1}$ $(13.6\text{ eV})$

Hint:

Always check whether the atom/ion is hydrogen-like (single electron). Use $Z^2$ scaling directly to compare transition energies quickly.

Answer 1

The Correct Option is C

To choose the correct energy of transition among the given options, we need to understand the basic concept of electronic transitions in hydrogen-like atoms. The energy of a transition from a higher energy level \(n_2\) to a lower energy level \(n_1\) can be calculated using the formula:

\(E = 13.6 \cdot Z^2 \cdot \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ eV}\)

where \(E\) is the energy of the transition, \(Z\) is the atomic number of the element, \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and higher energy levels, respectively.

1. Hydrogen (\(H\)) from \(n=2\) to \(n=1\):
   \(Z = 1\)
   \(E = 13.6 \cdot 1^2 \cdot \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \cdot \left(\frac{3}{4}\right) = 10.2 \text{ eV}\)
   This does not match the given \(6.8\text{ eV}\).
2. Lithium \((Li^{2+})\) from \(n=2\) to \(n=1\):
   \(Z = 3\)
   \(E = 13.6 \cdot 3^2 \cdot \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \cdot 9 \cdot \left(\frac{3}{4}\right) = 91.8 \text{ eV}\)
   This does not match the given \(13.6\text{ eV}\).
3. Helium \((He^+)\) from \(n=2\) to \(n=1\):
   \(Z = 2\)
   \(E = 13.6 \cdot 2^2 \cdot \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \cdot 4 \cdot \left(\frac{3}{4}\right) = 40.8 \text{ eV}\)
   This matches the given \(40.8\text{ eV}\) and is correct.
4. Beryllium \((Be^{3+})\) from \(n=2\) to \(n=1\):
   \(Z = 4\)
   \(E = 13.6 \cdot 4^2 \cdot \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6 \cdot 16 \cdot \left(\frac{3}{4}\right) = 163.2 \text{ eV}\)
   This does not match the given \(13.6\text{ eV}\).

Based on the calculations above, the correct energy of transition for the helium ion \(He^+\) from \(n=2\) to \(n=1\) is \(40.8\text{ eV}\). Therefore, the correct answer is:

$He^{+}_{2\to1}$ \(40.8\text{ eV}\).