The first excitation potential of a given atom is `10.2V`,then the ionisation potential is

Options

A. `10.2V`
B. `13.6V`
C. `30.6V`
D. `20.4V`

Answer

To find the ionization potential of an atom given its first excitation potential, we can follow these steps:

### Step 1: Understand the Energy Levels
The energy levels of an atom can be described using the formula:
\[ E_n = -\frac{E}{n^2} \]
where \(E\) is the ionization energy and \(n\) is the principal quantum number.

### Step 2: Identify the Ground State and First Excited State
For the ground state (n=1):
\[ E_1 = -E \]
For the first excited state (n=2):
\[ E_2 = -\frac{E}{2^2} = -\frac{E}{4} \]

### Step 3: Relate the Excitation Potential to Energy Levels
The first excitation potential is given as \(10.2V\). The difference in energy between the ground state and the first excited state is:
\[ E_1 - E_2 = 10.2V \]
Substituting the expressions for \(E_1\) and \(E_2\):
\[ -E - \left(-\frac{E}{4}\right) = 10.2 \]
This simplifies to:
\[ -E + \frac{E}{4} = 10.2 \]

### Step 4: Solve for Ionization Energy \(E\)
Rearranging the equation:
\[ -\frac{4E}{4} + \frac{E}{4} = 10.2 \]
\[ -\frac{3E}{4} = 10.2 \]
Multiplying both sides by -4/3:
\[ E = 10.2 \times \frac{4}{3} \]
Calculating this gives:
\[ E = 13.6V \]

### Conclusion
The ionization potential of the atom is:
\[ \text{Ionization Potential} = 13.6V \]

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