The acceptor level of a p-type semiconductor is $$6 \text{ eV}$$. The maximum wavelength of light which can create a hole would be : Given $$hc = 1242 \text{ eVnm}$$.

To create a hole in a p-type semiconductor, an electron must be excited from the valence band to the acceptor energy level. The minimum energy required for this transition is the energy gap of the acceptor level.

$$ E = 6 \text{ eV} $$

$$ E = \frac{hc}{\lambda} $$

$$ \lambda_{max} = \frac{hc}{E} $$

$$ hc = 1242 \text{ eVnm} $$ 

$$ E = 6 \text{ eV} $$

$$ \lambda_{max} = \frac{1242}{6} $$

$$ \lambda_{max} = 207 \text{ nm} $$