The ionisation potential of Hydrogen is `2.17 xx 10^(-11)` erg/atom. The energy of the electron in the second orbit of the hydrogen atom is

Options

A. `-(2.17 xx 10^(-11))/(2)`
B. `-(2.17 xx 10^(-11))/(2^(2))`
C. `-(2.17 xx 10^(17))/(2^(2))`
D. `-(2.17 xx 10^(11))/(2^(2))`

Answer

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

\[ E_n = -\frac{2.17 \times 10^{-11}}{n^2} \]

where:
- \( E_n \) is the energy of the electron in the nth orbit,
- \( n \) is the principal quantum number (orbit number),
- \( 2.17 \times 10^{-11} \) erg/atom is the ionization potential of hydrogen.

### Step 1: Identify the value of \( n \)
Since we are looking for the energy of the electron in the second orbit, we have:
\[ n = 2 \]

### Step 2: Substitute \( n \) into the energy formula
Now, we substitute \( n = 2 \) into the formula:

\[ E_2 = -\frac{2.17 \times 10^{-11}}{2^2} \]

### Step 3: Calculate \( 2^2 \)
Calculate \( 2^2 \):
\[ 2^2 = 4 \]

### Step 4: Substitute \( 2^2 \) into the energy formula
Now, substitute \( 4 \) back into the equation:
\[ E_2 = -\frac{2.17 \times 10^{-11}}{4} \]

### Step 5: Perform the division
Now, perform the division:
\[ E_2 = -\frac{2.17 \times 10^{-11}}{4} = -0.5425 \times 10^{-11} \]

### Step 6: Write the final answer
Thus, the energy of the electron in the second orbit of the hydrogen atom is:
\[ E_2 = -5.425 \times 10^{-12} \text{ erg/atom} \]

### Summary of the Solution:
The energy of the electron in the second orbit of the hydrogen atom is:
\[ E_2 = -5.425 \times 10^{-12} \text{ erg/atom} \]

---