A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in `Li^(++)` from the first to the third Bohr orbit (ionization energy of the hydrogen atom equals 13.6 eV).

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: Identify the parameters
For a hydrogen-like atom, the energy levels can be calculated using the formula:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

For Li²⁺, \( Z = 3 \).

### Step 2: Calculate the energy of the first and third orbits
- For the first orbit (\( n = 1 \)):
\[ E_1 = -\frac{3^2 \cdot 13.6}{1^2} = -\frac{9 \cdot 13.6}{1} = -122.4 \, \text{eV} \]

- For the third orbit (\( n = 3 \)):
\[ E_3 = -\frac{3^2 \cdot 13.6}{3^2} = -\frac{9 \cdot 13.6}{9} = -13.6 \, \text{eV} \]

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

### Step 4: Calculate the wavelength using the energy
The wavelength (\( \lambda \)) can be calculated using the formula:
\[ \lambda = \frac{hc}{\Delta E} \]
where \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \, \text{eV s} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

In terms of electron volts and angstroms, we can use:
\[ hc \approx 1240 \, \text{eV} \cdot \text{nm} \]
or
\[ hc \approx 12375 \, \text{eV} \cdot \text{Å} \]

Thus,
\[ \lambda = \frac{12375 \, \text{eV} \cdot \text{Å}}{108.8 \, \text{eV}} \approx 113.7 \, \text{Å} \]

### Final Answer
The wavelength of the radiation required to excite the electron in \( \text{Li}^{++} \) from the first to the third Bohr orbit is approximately **113.7 Å**.

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