Question

The acceptor level of a p-type semiconductor is \( 6 \) eV. 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

Hint:

Higher energy photons (lower wavelength) are required to excite electrons in semiconductors with a larger band gap.

Answer 1

The Correct Option is C

Step 1: {Using Energy-Wavelength Relation}
The energy of the photon required to excite an electron is given by: \[ E = \frac{hc}{\lambda} \] Rearranging for \( \lambda \): \[ \lambda = \frac{hc}{E} \] Step 2: {Substituting the Given Values}
\[ \lambda = \frac{1242}{6} \] \[ \lambda = 207 { nm} \] Thus, the correct answer is \( 207 \) nm.