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.
- 407 nm
- 414 nm
- 207 nm
- 103.5 nm
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to determine the maximum wavelength of light that can create a hole in a p-type semiconductor with an acceptor level of 6 eV. This involves using the relationship between energy, wavelength, and the quantum nature of light.
The energy \(E\) of a photon is related to its wavelength \(\lambda\) via the equation:
\(E = \frac{hc}{\lambda}\)
where:
- \(h\) is Planck's constant.
- \(c\) is the speed of light.
- \(\lambda\) is the wavelength.
Given that \(hc = 1242 \, \text{eV nm}\) and the acceptor level energy \(E = 6 \, \text{eV}\), we need to find the maximum wavelength \(\lambda\) that can induce transitions corresponding to this energy level.
Re-arranging the formula for wavelength, we get:
\(\lambda = \frac{hc}{E}\)
Substituting the provided values:
\(\lambda = \frac{1242 \, \text{eV nm}}{6 \, \text{eV}}\)
Calculating this gives:
\(\lambda = 207 \, \text{nm}\)
Thus, the maximum wavelength of light that can create a hole in this semiconductor is 207 nm.
Let's justify this answer:
- Option: 407 nm — This option would imply a lower energy than required.
- Option: 414 nm — This option also implies a lower energy than required.
- Option: 207 nm — Correct, as calculated.
- Option: 103.5 nm — This implies higher energy, which isn’t needed to create the hole.
Therefore, the correct answer is 207 nm.
Approach Solution -2
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 \( \lambda \):
\[\lambda = \frac{1240}{6} = 207 \, \text{nm}\]
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