An electron in a hydrogen atom in its ground state absorbs energy equal to ionisation energy of `Li^(+2)`. The wavelength of the emitted electron is :-

Options

A. `3.32xx10^(10)m`
B. `1.17 Å`
C. `2.32xx10^(9)nm`
D. `3.33"pm"`

Explanation

To solve the problem of finding the wavelength of the emitted electron when an electron in a hydrogen atom in its ground state absorbs energy equal to the ionization energy of \( \text{Li}^{2+} \), we can follow these steps:

### Step 1: Determine the Ionization Energy of \( \text{Li}^{2+} \)

The ionization energy for a hydrogen-like atom can be calculated using the formula:
\[ E = 13.6 \, \text{eV} \times Z^2 \]
where \( Z \) is the atomic number of the element. For lithium (\( \text{Li} \)), \( Z = 3 \).

So, the ionization energy of \( \text{Li}^{2+} \) is:
\[ E = 13.6 \, \text{eV} \times 3^2 = 13.6 \, \text{eV} \times 9 = 122.4 \, \text{eV} \]

### Step 2: Determine the Energy of the Hydrogen Atom in Ground State

The energy of an electron in the ground state of hydrogen (\( Z = 1 \)) is:
\[ E_{\text{H}} = 13.6 \, \text{eV} \]

### Step 3: Calculate the Energy of the Emitted Electron

When the electron in hydrogen absorbs energy equal to the ionization energy of \( \text{Li}^{2+} \), the energy absorbed is 122.4 eV. The energy of the emitted electron can be calculated as:
\[ E_{\text{emitted}} = E_{\text{absorbed}} - E_{\text{H}} = 122.4 \, \text{eV} - 13.6 \, \text{eV} = 108.8 \, \text{eV} \]

### Step 4: Convert the Energy of the Emitted Electron to Joules

To convert the energy from electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E_{\text{emitted}} = 108.8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.7408 \times 10^{-17} \, \text{J} \]

### Step 5: Calculate the Wavelength of the Emitted Electron

Using the formula relating energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), we can rearrange this to find \( \lambda \):
\[ \lambda = \frac{hc}{E} \]

Substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{1.7408 \times 10^{-17} \, \text{J}} \approx 1.144 \times 10^{-10} \, \text{m} \]

### Step 6: Convert Wavelength to Angstroms

To convert meters to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 1.144 \, \text{Å} \]

### Final Answer

The wavelength of the emitted electron is approximately:
\[ \lambda \approx 1.144 \, \text{Å} \]