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 to the third Bohr orbit (ionization energy of the hydrogen atom equals 13.6 eV).

Explanation

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

### Step 1: Determine the Energy of the Electron in the Initial and Final States
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 Li²⁺, \( Z = 3 \).

#### For the first orbit (n = 1):
\[ E_1 = -\frac{13.6 \times 3^2}{1^2} = -\frac{13.6 \times 9}{1} = -122.4 \, \text{eV} \]

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

### Step 2: Calculate the Energy Difference
The energy required to excite the electron from the first orbit to the third orbit is given by:

\[ \Delta E = E_3 - E_1 \]

Substituting the values we calculated:

\[ \Delta E = -13.6 - (-122.4) = -13.6 + 122.4 = 108.8 \, \text{eV} \]

### Step 3: Relate Energy to Wavelength
The energy of a photon is related to its wavelength by the equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J s}) \)
- \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \)

We can rearrange this equation to solve for the wavelength \( \lambda \):

\[ \lambda = \frac{hc}{E} \]

### Step 4: Convert Energy from eV to Joules
To use the above formula, we need to convert the energy from electron volts to joules. The conversion factor is:

\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

Thus,

\[ E = 108.8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.7408 \times 10^{-17} \, \text{J} \]

### Step 5: Calculate the Wavelength
Now we can substitute the values into the wavelength formula:

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{1.7408 \times 10^{-17} \, \text{J}} \]

Calculating this gives:

\[ \lambda \approx 1.14 \times 10^{-8} \, \text{m} = 114.25 \, \text{Å} \]

### 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 **114.25 Å**.

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