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 will follow these steps:

### Step 1: Determine the Energy Levels
The energy of an electron in the nth orbit of 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 lithium (Li²⁺), the atomic number \( Z = 3 \).

### Step 2: Calculate the Energy for the First Orbit (n=1)
Using the formula:
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -\frac{13.6 \times 9}{1} = -122.4 \text{ eV} \]

### Step 3: Calculate the Energy for the Third Orbit (n=3)
Using the formula:
\[ E_3 = -\frac{13.6 \times 3^2}{3^2} = -\frac{13.6 \times 9}{9} = -13.6 \text{ eV} \]

### Step 4: Calculate the Change in Energy (ΔE)
The change in energy when moving from the first to the third orbit is given by:
\[ \Delta E = E_3 - E_1 \]
Substituting the values:
\[ \Delta E = -13.6 - (-122.4) = -13.6 + 122.4 = 108.8 \text{ eV} \]

### Step 5: Convert ΔE to Joules
To convert the energy from 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 6: Calculate the Wavelength (λ)
Using the energy-wavelength relationship:
\[ \lambda = \frac{hc}{\Delta E} \]
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:
\[ \lambda = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{1.7408 \times 10^{-17} \text{ J}} \]

Calculating:
\[ \lambda = \frac{1.989 \times 10^{-25}}{1.7408 \times 10^{-17}} \approx 1.143 \times 10^{-8} \text{ m} = 114.3 \text{ nm} \]

### Step 7: Convert to Angstroms
Since \( 1 \text{ nm} = 10 \text{ Å} \):
\[ \lambda \approx 114.3 \text{ nm} = 1143 \text{ Å} \]

Thus, the wavelength of the radiation required to excite the electron in Li²⁺ from the first to the third Bohr orbit is approximately **1143 Å**.

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