Question

The acceptor level of a p-type semiconductor is 6eV. The maximum wavelength of light which can create a hole would be : Given hc = 1242 eV nm.

• A. 407 nm
• B. 414 nm
• C. 207 nm
• D. 103.5 nm

Answer 1

The Correct Option is C

The energy \( E \) of a photon is given by:
\[E = \frac{hc}{\lambda}\]
\[35\]
Substitute \( E = 6 \, \text{eV} \) and \( hc = 1240 \, \text{eV} \text{nm} \):
\[6 = \frac{1240}{\lambda \, (\text{nm})}\]
Rearrange to find \( \lambda \):
\[\lambda = \frac{1240}{6} = 207 \, \text{nm}\]