If the I excitation potential of a hypothetical H-like is 162 V, then the value of II excitation energy is about :

Options

A. 192 eV
B. 30 eV
C. 216 eV
D. 40.5 eV

Explanation

To solve the problem of finding the second excitation energy of a hypothetical hydrogen-like atom given that the first excitation potential is 162 V, we can follow these steps:

### Step 1: Understand the Formula for Excitation Potential
The excitation potential for the first excitation level of a hydrogen-like atom is given by the formula:
\[ E_1 = 10.2 \times Z^2 \]
where \(E_1\) is the first excitation potential and \(Z\) is the atomic number.

### Step 2: Substitute the Given Value
We know from the problem that \(E_1 = 162 \, \text{V}\). We can set up the equation:
\[ 162 = 10.2 \times Z^2 \]

### Step 3: Solve for \(Z^2\)
To find \(Z^2\), we rearrange the equation:
\[ Z^2 = \frac{162}{10.2} \]
Calculating this gives:
\[ Z^2 = 15.8824 \]

### Step 4: Use the Formula for Second Excitation Energy
The formula for the second excitation energy \(E_2\) for hydrogen-like species is:
\[ E_2 = 12.09 \times Z^2 \]

### Step 5: Substitute \(Z^2\) into the Second Excitation Energy Formula
Now, we substitute the value of \(Z^2\) into the equation:
\[ E_2 = 12.09 \times 15.8824 \]

### Step 6: Calculate \(E_2\)
Performing the multiplication:
\[ E_2 \approx 192.01 \, \text{eV} \]

### Step 7: Round to the Nearest Whole Number
The nearest whole number to 192.01 is:
\[ E_2 \approx 192 \, \text{eV} \]

### Conclusion
Thus, the value of the second excitation energy is approximately **192 eV**.