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 problem requires determining the maximum light wavelength capable of generating a hole in a p-type semiconductor with an acceptor level at 6 eV. This is achieved by applying the energy-wavelength relationship for photons.

The energy \(E\) of a photon is defined by its wavelength \(\lambda\) through the equation:

\(E = \frac{hc}{\lambda}\)

Where:

• \(h\) represents Planck's constant.
• \(c\) denotes the speed of light.
• \(\lambda\) is the wavelength.

Given the constant \(hc = 1242 \, \text{eV nm}\) and the acceptor level energy \(E = 6 \, \text{eV}\), we aim to find the maximum wavelength \(\lambda\) that facilitates energy transitions to this level.

Rearranging the equation to solve for wavelength yields:

\(\lambda = \frac{hc}{E}\)

Substituting the given values:

\(\lambda = \frac{1242 \, \text{eV nm}}{6 \, \text{eV}}\)

The calculation results in:

\(\lambda = 207 \, \text{nm}\)

Consequently, the maximum wavelength of light that can create a hole in this semiconductor is 207 nm.

Justification of the answer:

• Option: 407 nm — Corresponds to energy below the required 6 eV.
• Option: 414 nm — Also corresponds to energy below the required 6 eV.
• Option: 207 nm — Matches the calculated value.
• Option: 103.5 nm — Implies energy higher than necessary for creating the hole.

Therefore, the correct answer is 207 nm.