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.

Answer 2

Step 1: Understanding the Physical Concept
In a p-type semiconductor, the acceptor level lies just above the valence band. When light shines on the material, photons can excite electrons from the valence band into the acceptor level, effectively leaving behind holes (which are positive charge carriers in semiconductors). To do this, the photon must have **at least** the energy equal to the gap between the valence band and the acceptor level.

Step 2: Given Data
From the question:

The energy required to excite an electron (and hence create a hole) is \( E = 6 \, \text{eV} \).

The constant\(hc = 1242 \, {eV·nm}\)This is the product of Planck's constant and the speed of light, converted to units that make calculating wavelength straightforward when energy is in eV and wavelength is in nm.

 

Step 3: Formula Used
We use the energy-wavelength relation of photons: \[ E = \frac{hc}{\lambda} \] Solving for wavelength \( \lambda \), we get: \[ \lambda = \frac{hc}{E} \]

Step 4: Substituting the Values
Now substitute the given values into the formula:

\[\lambda = \frac{1242 \,{eV·nm}}{6 \, \text{eV}} = 207 \, \text{nm}\]

Step 5: Interpreting the Result
The calculated wavelength of 207 nm represents the maximum wavelength of light that still has enough energy to create a hole in the p-type semiconductor. Any wavelength longer than this (i.e., lower energy) won't have enough energy to excite the electron and thus cannot generate a hole.

Step 6: Final Answer
\[ \boxed{\lambda = 207 \, \text{nm}} \] Therefore, the correct answer is: Option 3: 207 nm.

Answer 3

Step 1: Understanding the Physical Concept
In a p-type semiconductor, the acceptor level lies just above the valence band. When light shines on the material, photons can excite electrons from the valence band into the acceptor level, effectively leaving behind holes (which are positive charge carriers in semiconductors). To do this, the photon must have **at least** the energy equal to the gap between the valence band and the acceptor level.

Step 2: Given Data
From the question:

The energy required to excite an electron (and hence create a hole) is \( E = 6 \, \text{eV} \).

The constant\(hc = 1242 \, {eV·nm}\)This is the product of Planck's constant and the speed of light, converted to units that make calculating wavelength straightforward when energy is in eV and wavelength is in nm.

 

Step 3: Formula Used
We use the energy-wavelength relation of photons: \[ E = \frac{hc}{\lambda} \] Solving for wavelength \( \lambda \), we get: \[ \lambda = \frac{hc}{E} \]

Step 4: Substituting the Values
Now substitute the given values into the formula:

\[\lambda = \frac{1242 \,{eV·nm}}{6 \, \text{eV}} = 207 \, \text{nm}\]

Step 5: Interpreting the Result
The calculated wavelength of 207 nm represents the maximum wavelength of light that still has enough energy to create a hole in the p-type semiconductor. Any wavelength longer than this (i.e., lower energy) won't have enough energy to excite the electron and thus cannot generate a hole.

Step 6: Final Answer
\[ \boxed{\lambda = 207 \, \text{nm}} \] Therefore, the correct answer is: Option 3: 207 nm.