The ratio of energy of the electron in ground state of hydrogen to the electron in first excited state of `Be^(3+)` is

Options

A. ` 1 : 4`
B. `1 : 8`
C. `1 : 16`
D. `16 : 1`

Answer

To solve the problem of finding the ratio of the energy of the electron in the ground state of hydrogen to the electron in the first excited state of \( \text{Be}^{3+} \), we will use the formula for the energy levels of hydrogen-like atoms:

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

### Step 1: Calculate the energy of the electron in the ground state of hydrogen

For hydrogen (\( Z = 1 \)) in the ground state (\( n = 1 \)):
\[ E_1 = -\frac{13.6 \cdot 1^2}{1^2} = -13.6 \, \text{eV} \]

### Step 2: Calculate the energy of the electron in the first excited state of \( \text{Be}^{3+} \)

For \( \text{Be}^{3+} \) (\( Z = 4 \)) in the first excited state (\( n = 2 \)):
\[ E_2 = -\frac{13.6 \cdot 4^2}{2^2} = -\frac{13.6 \cdot 16}{4} = -\frac{217.6}{4} = -54.4 \, \text{eV} \]

### Step 3: Calculate the ratio of the energies

Now, we need to find the ratio \( \frac{E_1}{E_2} \):
\[ \frac{E_1}{E_2} = \frac{-13.6}{-54.4} = \frac{13.6}{54.4} \]

### Step 4: Simplify the ratio

To simplify \( \frac{13.6}{54.4} \):
\[ \frac{13.6}{54.4} = \frac{1}{4} \]

### Conclusion

Thus, the ratio of the energy of the electron in the ground state of hydrogen to the energy of the electron in the first excited state of \( \text{Be}^{3+} \) is:
\[ \text{Ratio} = 1 : 4 \]

### Final Answer
The correct option is \( 1 : 4 \).

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