To find the wavelength of the emitted electron, we need to first find the energy absorbed by the electron in the hydrogen atom and then use that energy to calculate the wavelength of the emitted electron.

The ionization energy of Li 2 is the energy required to remove an electron from the Li 2 atom. Let's assume the ionization energy of Li 2 is E.

Now, when the electron in the hydrogen atom absorbs energy equal to the ionization energy of Li 2, it gets excited to a higher energy level. This means that the electron is now in an excited state.

To find the energy absorbed by the electron in the hydrogen atom, we need to find the difference in energy between the ground state and the excited state.

The energy of an electron in the ground state of a hydrogen atom is given by the formula:

E1 = -13.6 eV

The energy of an electron in an excited state of a hydrogen atom is given by the formula:

E2 = -13.6 eV / n^2

where n is the principal quantum number of the excited state.

Since the electron absorbed energy equal to the ionization energy of Li 2, we can equate E2 to E:

E2 = E

-13.6 eV / n^2 = E

Now, let's solve for n:

n^2 = -13.6 eV / E

n = sqrt(-13.6 eV / E)

Since we don't know the exact value of E, we cannot calculate the exact value of n. However, we know that the electron is in a higher energy level than the ground state, so n must be greater than 1.

Now, the wavelength of the emitted electron can be calculated using the equation:

wavelength = h / (E2 - E1)

where h is the Planck's constant.

Since we know the values of E1 and E2, we can substitute them into the equation:

wavelength = h / (-13.6 eV / n^2 - (-13.6 eV))

wavelength = h / (-13.6 eV / n^2 + 13.6 eV)

wavelength = h / (13.6 eV - 13.6 eV / n^2)

Since we don't know the exact value of n, we cannot calculate the exact value of the wavelength. However, we know that n is greater than 1, so the wavelength must be greater than 3.32. Therefore, the correct answer is:

b) greater than 3.32