A doubly ionized lithium atom is hydrogen like with atomic
. number 3. Find the wavelength of the radiation to excite the electron in
. `Li++` form the first to the third Bohr orbit. The ionization energy of the hydrogen
. Atom is 13.6V.

Explanation

To find the wavelength of the radiation required to excite the electron in a doubly ionized lithium atom (Li²⁺) from the first to the third Bohr orbit, we can follow these steps:

### Step 1: Determine the energy levels using the Bohr model
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

For Li²⁺, \( Z = 3 \).

### Step 2: Calculate the energy for the first and third orbits
1. **Energy at \( n = 1 \)**:
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -\frac{13.6 \times 9}{1} = -122.4 \, \text{eV} \]

2. **Energy at \( n = 3 \)**:
\[ E_3 = -\frac{13.6 \times 3^2}{3^2} = -\frac{13.6 \times 9}{9} = -13.6 \, \text{eV} \]

### Step 3: Calculate the change in energy (\( \Delta E \))
The change in energy when the electron moves from the first orbit to the third orbit is:
\[ \Delta E = E_3 - E_1 = -13.6 - (-122.4) = -13.6 + 122.4 = 108.8 \, \text{eV} \]

### Step 4: Convert the energy change to Joules
To convert electron volts to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = 108.8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.7408 \times 10^{-17} \, \text{J} \]

### Step 5: Calculate the wavelength using the energy-wavelength relation
The wavelength \( \lambda \) can be calculated using the formula:
\[ \lambda = \frac{hc}{\Delta E} \]
where \( h = 6.63 \times 10^{-34} \, \text{J s} \) (Planck's constant) and \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light).

Substituting the values:
\[ \lambda = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{1.7408 \times 10^{-17}} \]

### Step 6: Calculate the wavelength
Calculating the above expression:
\[ \lambda = \frac{1.989 \times 10^{-25}}{1.7408 \times 10^{-17}} \approx 1.143 \times 10^{-8} \, \text{m} = 114.3 \, \text{nm} \]

### Final Answer
The wavelength of the radiation required to excite the electron in \( \text{Li}^{2+} \) from the first to the third Bohr orbit is approximately **114.3 nm**.

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