The energy of an electron in the first Bohr's orbit of a hydrogen atom is `-2.18 xx10^(-18)J`. Its energy in the second orbit would be 

Options

A. `-1.09 xx 10^(-18)J`
B. `-4.36 xx 10^(-18)J`
C. `-5.45 xx 10^(-19)J`
D. `-8.72 xx 10^(-18)J`

Answer

To find the energy of an electron in the second Bohr orbit of a hydrogen atom, we can use the formula for the energy of an electron in the nth orbit:

\[ E_n = -\frac{2.18 \times 10^{-18} \text{ J}}{n^2} \]

where:
- \( E_n \) is the energy of the electron in the nth orbit,
- \( n \) is the principal quantum number (1 for the first orbit, 2 for the second orbit), and
- \( 2.18 \times 10^{-18} \text{ J} \) is the energy of the electron in the first orbit (n=1).

### Step-by-Step Solution:

1. **Identify the given values:**
- Energy in the first orbit \( E_1 = -2.18 \times 10^{-18} \text{ J} \)
- Principal quantum number for the second orbit \( n = 2 \)

2. **Use the formula for energy in the nth orbit:**
\[ E_n = -\frac{2.18 \times 10^{-18} \text{ J}}{n^2} \]

3. **Substitute \( n = 2 \) into the formula:**
\[ E_2 = -\frac{2.18 \times 10^{-18} \text{ J}}{2^2} \]

4. **Calculate \( 2^2 \):**
\[ 2^2 = 4 \]

5. **Now substitute this value back into the equation:**
\[ E_2 = -\frac{2.18 \times 10^{-18} \text{ J}}{4} \]

6. **Perform the division:**
\[ E_2 = -0.545 \times 10^{-18} \text{ J} \]

7. **Express the result in scientific notation:**
\[ E_2 = -5.45 \times 10^{-19} \text{ J} \]

### Final Answer:
The energy of the electron in the second orbit of a hydrogen atom is \( -5.45 \times 10^{-19} \text{ J} \).

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