Question

A sample of hydrogen atom in its ground state is radiated with photons of $10.2$ eV energies. The radiation from the sample is absorbed by excited ionized He+. Then which of the following statement(s) is/are true?

• A. He+ electron moves from n = 2 to n = 4
• B. In the He+ emission spectra, there will be 6 lines.
• C. Smallest wavelength of He+ spectrum is obtained when transition taken place from n=4 to n=3
• D. He+ electron moves from n=2 to n=3.

Answer 1

The Correct Option is A, B

First, we need to analyze the energy levels of the hydrogen atom and the He⁺ ion, and then determine the possible transitions that can occur.

Step 1: Energy Levels of Hydrogen

The energy of an electron in the \(n\)-th energy level of a hydrogen atom is given by: \[E_n = - \frac{13.6}{n^2} \text{ eV}\]

The ground state energy of hydrogen (\(n = 1\)) is \(E_1 = -13.6\) eV. When hydrogen atoms absorb photons of \(10.2\) eV, the electron transitions from \(n = 1\) to a higher energy level. The energy difference is: \[\Delta E = E_n - E_1 = 10.2 \text{ eV}\] \[E_n = E_1 + 10.2 = -13.6 + 10.2 = -3.4 \text{ eV}\] \[-3.4 = - \frac{13.6}{n^2} \Rightarrow n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2\]

So, the hydrogen atom transitions from \(n = 1\) to \(n = 2\). The emitted photons will have \(10.2\) eV energy.

Step 2: Energy Levels of He⁺

The energy of an electron in the \(n\)-th energy level of a He⁺ ion (which has atomic number \(Z = 2\)) is given by: \[E_n = - \frac{Z^2 \times 13.6}{n^2} = - \frac{4 \times 13.6}{n^2} = - \frac{54.4}{n^2} \text{ eV}\]

Step 3: Possible Transitions in He⁺

We need to find an initial and final energy level in He⁺ such that the energy difference is \(10.2\) eV. Let's assume the He⁺ electron is initially in the \(n_i\) level and transitions to the \(n_f\) level. \[\Delta E = E_{n_f} - E_{n_i} = - \frac{54.4}{n_f^2} - \left( - \frac{54.4}{n_i^2} \right) = 54.4 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) = 10.2\] \[\frac{1}{n_i^2} - \frac{1}{n_f^2} = \frac{10.2}{54.4} = \frac{1}{5.33} \approx \frac{3}{16}\] If we assume the electron is initially in n=2, \[\frac{1}{4} - \frac{1}{n_f^2} = \frac{3}{16} \implies \frac{1}{n_f^2} = \frac{1}{4} - \frac{3}{16} = \frac{4-3}{16} = \frac{1}{16}\] This implies \(n_f^2 = 16\) or \(n_f = 4\) The He+ electron does move from n=2 to n=4.

Step 4: Number of Emission Lines

If the electron transitions from \(n = 4\) to the ground state, the possible transitions are:

• 4 → 3
• 4 → 2
• 4 → 1
• 3 → 2
• 3 → 1
• 2 → 1

Therefore, there will be 6 lines in the emission spectra.

 

Step 5: Smallest Wavelength

The smallest wavelength corresponds to the transition with the largest energy difference. In this case, it is not from 4 to 3 but either 4 to 1, 3 to 1 or 2 to 1.

Step 6: Determine the Correct Statements

1. He⁺ electron moves from n = 2 to n = 4: This statement is TRUE.
2. In the He⁺ emission spectra, there will be 6 lines: This statement is TRUE.
3. Smallest wavelength of He⁺ spectrum is obtained when transition taken place from n=4 to n=3: This statement is FALSE.
4. He⁺ electron moves from n=2 to n=3: This statement is FALSE.

Conclusion

The correct statements are:

• He⁺ electron moves from n = 2 to n = 4
• In the He⁺ emission spectra, there will be 6 lines.