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: {Apply the Energy-Wavelength Relationship}
The photon energy needed for electron excitation is calculated using:\[E = \frac{hc}{\lambda}\]To find the wavelength \( \lambda \):\[\lambda = \frac{hc}{E}\]Step 2: {Input Provided Values}
\[\lambda = \frac{1242}{6}\]\[\lambda = 207 { nm}\]The resultant wavelength is \( 207 \) nm.