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
Approach Solution - 1
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:
- 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 (\(n = 1, 2, 3, \ldots\)).
- 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.
Approach Solution -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:
\[ \text{No. of transitions} = \frac{n(n - 1)}{2} = \frac{4 \times (4 - 1)}{2} = 6 \]
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