The ratio of ionization energy of `H` and `Be^(+3)` is.

Options

A. `1:1`
B. `1:3`
C. `1:9`
D. `1:16`

Explanation

To find the ratio of the ionization energy of hydrogen (H) to that of beryllium ion (Be^(+3)), we can follow these steps:

### Step 1: Understand Ionization Energy
The ionization energy (IE) is the energy required to remove an electron from an atom in its gaseous state. For hydrogen, the ionization energy can be calculated using the formula:
\[ \text{IE} = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number of the electron being removed.

### Step 2: Calculate Ionization Energy of Hydrogen
For hydrogen (H):
- Atomic number \( Z = 1 \)
- The electron is in the ground state, so \( n = 1 \).

Using the formula:
\[ \text{IE}_{H} = \frac{13.6 \, \text{eV} \cdot (1)^2}{(1)^2} = 13.6 \, \text{eV} \]

### Step 3: Calculate Ionization Energy of Beryllium Ion (Be^(+3))
For beryllium ion (Be^(+3)):
- Atomic number \( Z = 4 \)
- The ion has lost 3 electrons, leaving one electron in the ground state, so \( n = 1 \).

Using the formula:
\[ \text{IE}_{Be^{+3}} = \frac{13.6 \, \text{eV} \cdot (4)^2}{(1)^2} = \frac{13.6 \, \text{eV} \cdot 16}{1} = 217.6 \, \text{eV} \]

### Step 4: Calculate the Ratio of Ionization Energies
Now, we can find the ratio of the ionization energy of hydrogen to that of beryllium ion:
\[ \text{Ratio} = \frac{\text{IE}_{H}}{\text{IE}_{Be^{+3}}} = \frac{13.6 \, \text{eV}}{217.6 \, \text{eV}} \]

### Step 5: Simplify the Ratio
To simplify:
\[ \text{Ratio} = \frac{13.6}{217.6} = \frac{1}{16} \]

### Conclusion
The ratio of the ionization energy of hydrogen to that of beryllium ion (Be^(+3)) is:
\[ \text{Ratio} = 1 : 16 \]