Askiitians Tutor Team
Last Activity: 4 Days ago
To determine the maximum wavelength of light that can create a hole in a p-type semiconductor, we first need to understand the relationship between energy and wavelength. The energy required to create a hole corresponds to the energy difference between the valence band and the acceptor level. In this case, the acceptor level is 57 meV above the valence band.
Energy-Wavelength Relationship
The energy of a photon can be calculated using the equation:
E = h * f
Where:
- E is the energy of the photon in joules.
- h is Planck's constant (approximately 6.626 x 10-34 J·s).
- f is the frequency of the photon in hertz.
We can also relate frequency to wavelength using the equation:
c = f * λ
Where:
- c is the speed of light (approximately 3.00 x 108 m/s).
- λ is the wavelength in meters.
Calculating the Energy in Joules
First, we need to convert the energy from millielectronvolts (meV) to joules. The conversion factor is:
1 eV = 1.602 x 10-19 J
Thus, to convert 57 meV to joules:
E = 57 meV * (1.602 x 10-19 J/eV) * (10-3 eV/meV)
E = 57 x 1.602 x 10-22 J
E ≈ 9.13 x 10-15 J
Finding the Frequency
Now that we have the energy in joules, we can find the frequency:
f = E / h
f = (9.13 x 10-15 J) / (6.626 x 10-34 J·s)
f ≈ 1.38 x 1019 Hz
Calculating the Wavelength
Finally, we can find the wavelength using the speed of light equation:
λ = c / f
λ = (3.00 x 108 m/s) / (1.38 x 1019 Hz)
λ ≈ 2.17 x 10-11 m
To express this in nanometers (1 nm = 10-9 m):
λ ≈ 0.0217 nm
Summary
The maximum wavelength of light that can create a hole in the p-type semiconductor, given that the acceptor level is 57 meV above the valence band, is approximately 0.0217 nm. This wavelength falls within the high-energy range of the electromagnetic spectrum, indicating that only very high-energy photons (like X-rays or gamma rays) can excite electrons from the valence band to the acceptor level, thereby creating holes.