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

Options

A. `1//4`
B. `4//9`
C. `9//4`
D. `4`

Explanation

To find the ratio of the energies of the hydrogen atom in its first and second excited states, we can use the formula for the energy of an electron in the nth orbit of a hydrogen atom, which is given by:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step 1: Identify the energy levels for the excited states
The first excited state corresponds to \(n = 2\) and the second excited state corresponds to \(n = 3\).

### Step 2: Calculate the energy for the first excited state (\(n = 2\))
Using the formula:

\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 3: Calculate the energy for the second excited state (\(n = 3\))
Using the formula:

\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

### Step 4: Find the ratio of the energies
Now, we can find the ratio of the energies \(E_2\) to \(E_3\):

\[ \frac{E_2}{E_3} = \frac{-3.4 \, \text{eV}}{-1.51 \, \text{eV}} = \frac{3.4}{1.51} \]

Calculating this gives:

\[ \frac{3.4}{1.51} \approx 2.25 \]

### Step 5: Express the ratio in a simpler form
To express the ratio in terms of whole numbers, we can multiply both the numerator and denominator by 100 to avoid decimals:

\[ \frac{340}{151} \approx \frac{9}{4} \]

### Conclusion
Thus, the ratio of the energies of the hydrogen atom in its first to second excited state is:

\[ \frac{E_2}{E_3} = \frac{9}{4} \]