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, we need to find 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 \( \text{Li}^{2+} \).

### Step-by-Step Solution:

**Step 1: Determine the Ionization Energy of \( \text{Li}^{2+} \)**
The ionization energy for a hydrogen-like atom is given by the formula:
\[ E = 13.6 \, \text{eV} \times Z^2 \]
where \( Z \) is the atomic number. For lithium (\( \text{Li} \)), \( Z = 3 \).
Thus, the ionization energy of \( \text{Li}^{2+} \) is:
\[ E_{\text{Li}^{2+}} = 13.6 \, \text{eV} \times 3^2 = 13.6 \, \text{eV} \times 9 = 122.4 \, \text{eV} \]

**Step 2: Determine the Energy of the Electron in Hydrogen**
The energy of the electron in the ground state of hydrogen is:
\[ E_{\text{H}} = -13.6 \, \text{eV} \]
Since we are considering the energy absorbed, we take the absolute value.

**Step 3: Calculate the Total Energy of the Emitted Electron**
When the electron absorbs energy equal to the ionization energy of \( \text{Li}^{2+} \), the total energy of the emitted electron is:
\[ E_{\text{emitted}} = E_{\text{Li}^{2+}} - E_{\text{H}} \]
Substituting the values:
\[ E_{\text{emitted}} = 122.4 \, \text{eV} - (-13.6 \, \text{eV}) = 122.4 \, \text{eV} + 13.6 \, \text{eV} = 136 \, \text{eV} \]

**Step 4: Calculate the Wavelength of the Emitted Electron**
The wavelength \( \lambda \) of the emitted electron can be calculated using the formula:
\[ \lambda = \frac{hc}{E} \]
where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( E = 136 \, \text{eV} \) (energy of the emitted electron)
To convert \( E \) from eV to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = 136 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.176 \times 10^{-17} \, \text{J} \]

Now substituting the values into the wavelength formula:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{2.176 \times 10^{-17} \, \text{J}} \]
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{2.176 \times 10^{-17} \, \text{J}} \]
\[ \lambda \approx 9.13 \times 10^{-9} \, \text{m} \]
To convert meters to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 91.3 \, \text{angstroms} \]

### Final Answer:
The wavelength of the emitted electron is approximately \( 91.3 \, \text{angstroms} \).