If the ionization energy of `He^(+)` is `19.6xx10^(-18) J` per atom then the energy of `Be^(3+)` ion in the second stationary state is `:`

Options

A. `-4.9xx10^(-18)J`
B. `-44.1xx10^(-18)J`
C. `-11.025xx10^(-18)J`
D. `-19.4xx10^(-18)J`

Explanation

To solve the problem of finding the energy of the `Be^(3+)` ion in the second stationary state given the ionization energy of `He^(+)`, we can follow these steps:

### Step 1: Understand the given information
We know that the ionization energy of `He^(+)` is `19.6 x 10^(-18) J`. We need to find the energy of `Be^(3+)` in the second stationary state.

### Step 2: Recall the formula for ionization energy
The ionization energy (IE) for a hydrogen-like atom can be expressed as:
\[ \text{IE} = -\frac{Z^2 \cdot R_H}{n^2} \]
where:
- \( Z \) is the atomic number,
- \( R_H \) is the Rydberg constant for hydrogen (approximately \( 2.18 \times 10^{-18} \, J \)),
- \( n \) is the principal quantum number (stationary state).

### Step 3: Calculate the ionization energy of hydrogen
From the ionization energy of `He^(+)`, we can find the ionization energy of hydrogen. Since for `He^(+)`, \( Z = 2 \) and \( n = 1 \):
\[ \text{IE}_{He^{+}} = -\frac{2^2 \cdot R_H}{1^2} \]
Given that:
\[ \text{IE}_{He^{+}} = 19.6 \times 10^{-18} \, J \]
We can set up the equation:
\[ 19.6 \times 10^{-18} = 4 \cdot R_H \]
Thus:
\[ R_H = \frac{19.6 \times 10^{-18}}{4} = 4.9 \times 10^{-18} \, J \]

### Step 4: Calculate the energy of `Be^(3+)` in the second stationary state
For `Be^(3+)`, \( Z = 4 \) and \( n = 2 \):
\[ \text{IE}_{Be^{3+}} = -\frac{4^2 \cdot R_H}{2^2} \]
Substituting the value of \( R_H \):
\[ \text{IE}_{Be^{3+}} = -\frac{16 \cdot 4.9 \times 10^{-18}}{4} \]
\[ \text{IE}_{Be^{3+}} = -16 \times 4.9 \times 10^{-18} \times \frac{1}{4} \]
\[ \text{IE}_{Be^{3+}} = -16 \times 1.225 \times 10^{-18} \]
\[ \text{IE}_{Be^{3+}} = -19.6 \times 10^{-18} \, J \]

### Step 5: Final result
The energy of the `Be^(3+)` ion in the second stationary state is:
\[ \text{Energy of } Be^{3+} = -19.6 \times 10^{-18} \, J \]