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 levels
The energy levels of a hydrogen atom are defined by the principal quantum number \( n \). The first excited state corresponds to \( n = 2 \) and the third excited state corresponds to \( n = 4 \).

### Step 2: Write the formula for energy levels
The energy of an electron in a hydrogen atom at a given energy level \( n \) is given by the formula:
\[ E_n = -\frac{Z^2 \cdot k}{n^2} \]
where:
- \( E_n \) is the energy at level \( n \),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( k \) is a constant.

### Step 3: Calculate the energies for the first and third excited states
For the first excited state (\( n = 2 \)):
\[ E_2 = -\frac{1^2 \cdot k}{2^2} = -\frac{k}{4} \]

For the third excited state (\( n = 4 \)):
\[ E_4 = -\frac{1^2 \cdot k}{4^2} = -\frac{k}{16} \]

### Step 4: Find the ratio of energies
Now, we can find the ratio of the energy in the first excited state to the energy in the third excited state:
\[ \text{Ratio} = \frac{E_2}{E_4} = \frac{-\frac{k}{4}}{-\frac{k}{16}} = \frac{\frac{k}{4}}{\frac{k}{16}} = \frac{16}{4} = 4 \]

### Final Answer
The ratio of energies of the hydrogen atom in its first excited state to its third excited state is \( 4 \).

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