Question

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 \( \ldots \).

Answer 1

Correct Answer: 6

The problem provides the energy of an electron in an excited state of a hydrogen atom as \( E_n = -0.85 \, \text{eV} \) and asks for the maximum number of possible transitions to lower energy levels.

Concept Used:

The solution involves two key concepts from the Bohr model of the hydrogen atom:

1. Energy Levels of Hydrogen Atom: The energy of an electron in the \(n\)-th principal energy level (or orbit) is given by the formula: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] where \(n\) is the principal quantum number (\(n = 1, 2, 3, \ldots\)).
2. Number of Spectral Lines: When an electron de-excites from a higher energy level \(n\) to lower energy levels, it can do so in a series of jumps. The maximum number of distinct spectral lines (or allowed transitions) that can be emitted is given by the formula: \[ \text{Number of transitions} = \frac{n(n-1)}{2} \]

Step-by-Step Solution:

Step 1: Determine the principal quantum number (\(n\)) of the excited state.

We are given the energy of the electron in the excited state, \(E_n = -0.85 \, \text{eV}\). We can use the energy level formula for the hydrogen atom to find the value of \(n\).

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

Now, we solve for \(n^2\):

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

To simplify the fraction, we can multiply the numerator and denominator by 100:

\[ n^2 = \frac{1360}{85} = \frac{272}{17} = 16 \]

Taking the square root of both sides to find \(n\):

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

This means the electron is in the 4th energy level, which is the 3rd excited state.

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

Now that we know the electron is in the \(n=4\) state, we can use the formula for the maximum number of spectral lines to find the total number of possible transitions to all lower energy levels (n=3, n=2, and n=1).

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

Substitute \(n=4\) into the formula:

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

Final Computation & Result:

Performing the final calculation:

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

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

Answer 2

Step 1. Calculate Quantum Number n: Use the energy formula for hydrogen:

\( E_n = -\frac{13.6}{n^2} = -0.85 \)

Solving, \( n = 4 \).

Step 2. Determine Number of Transitions: The number of transitions from \( n = 4 \) is:

\[ \text{No. of transitions} = \frac{n(n - 1)}{2} = \frac{4 \times (4 - 1)}{2} = 6 \]