The ratio of energies of hydrogen atom in its first excited state to third excited state is

Options

A. `1/4`
B. `4/1`
C. `3/4`
D. `4/3`

Answer

To find the ratio of energies of a hydrogen atom in its first excited state to its third excited state, we can follow these steps:

### Step 1: Identify the energy formula
The energy of an electron in a hydrogen atom in a specific energy level (n) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Determine the energy levels
For the first excited state, \( n = 2 \).
For the third excited state, \( n = 4 \).

### Step 3: Calculate the energies
Using the formula for energy, we can calculate the energies for both states.

1. **Energy in the first excited state (n = 2)**:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

2. **Energy in the third excited state (n = 4)**:
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

### Step 4: Find the ratio of energies
Now, we need to find the ratio of the energy in the first excited state to the energy in the third excited state:
\[ \text{Ratio} = \frac{E_2}{E_4} = \frac{-3.4 \, \text{eV}}{-0.85 \, \text{eV}} = \frac{3.4}{0.85} \]

### Step 5: Simplify the ratio
Calculating the ratio:
\[ \frac{3.4}{0.85} = 4 \]
Thus, the ratio of energies \( E_2 : E_4 = 4 : 1 \).

### Final Answer
The ratio of energies of the hydrogen atom in its first excited state to its third excited state is:
\[ \text{Ratio} = 4 : 1 \]

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