An electron of a hydrogen atom in an excited state is having energy $ E_n = -0.85 \, \text{eV} $. The maximum number of allowed transitions to lower energy levels is $ \ldots $.

Question

The wavelength of spectral line obtained in the spectrum of Li$^{2+}$ ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

Answer 1

Given the energy of an electron in an excited hydrogen atom as $ E_n = -0.85 \, \text{eV} $, determine the maximum number of possible transitions to lower energy levels.

Key Concepts:

This problem utilizes two fundamental principles of the Bohr model for hydrogen atoms:

Hydrogen Atom Energy Levels: The energy of an electron in the principal energy level $n$ is defined by: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] where $n$ represents the principal quantum number ($n = 1, 2, 3, \ldots$).
Number of Spectral Lines: When an electron transitions from a higher energy level to a lower one, it can emit distinct spectral lines. The maximum number of such transitions from level $n$ is calculated as: \[ \text{Number of transitions} = \frac{n(n-1)}{2} \]

Solution Steps:

Step 1: Identify the principal quantum number ($n$) of the excited state.

Using the provided energy $E_n = -0.85 \, \text{eV}$ and the hydrogen atom's energy level formula, we solve for $n$:

\[ -0.85 = -\frac{13.6}{n^2} \]

Rearranging to find $n^2$:

\[ n^2 = \frac{13.6}{0.85} \]

Simplifying the fraction:

\[ n^2 = \frac{1360}{85} = 16 \]

Taking the square root to find $n$:

\[ n = \sqrt{16} = 4 \]

The electron is in the 4th energy level, corresponding to the 3rd excited state.

Step 2: Compute the maximum number of allowed transitions from this level.

With $n=4$, we apply the formula for the maximum number of spectral lines to find all possible transitions to lower levels (n=3, n=2, and n=1).

\[ \text{Number of transitions} = \frac{n(n-1)}{2} \]

Substituting $n=4$:

\[ \text{Number of transitions} = \frac{4(4-1)}{2} \]

Final Calculation and Result:

The calculation yields:

\[ \text{Number of transitions} = \frac{4 \times 3}{2} = \frac{12}{2} = 6 \]

The maximum number of possible transitions to lower energy levels is 6.

An electron of a hydrogen atom in an excited state is having energy \( E_n = -0.85 \, \text{eV} \). The maximum number of allowed transitions to lower energy levels is \( \ldots \).

Correct Answer: 6

Solution and Explanation

Key Concepts:

Solution Steps:

Final Calculation and Result:

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