Given:
Acceptor level = 50 meV above valence band
1 eV = 1.6 × 10⁻¹⁹ J
h = 6.63 × 10⁻³⁴ J·s
c = 3 × 10⁸ m/s
Energy required to create a hole:
\[ E = 50 \text{ meV} = 50 \times 10^{-3} \times 1.6 \times 10^{-19} \text{ J} \]
\[ E = 8 \times 10^{-21} \text{ J} \]
Maximum wavelength (λₘₐₓ):
\[ \lambda_{max} = \frac{hc}{E} \]
\[ \lambda_{max} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{8 \times 10^{-21}} \]
\[ \lambda_{max} = 2.486 \times 10^{-5} \text{ m} \]
\[ \lambda_{max} \approx 2.48 \times 10^{-5} \text{ m} \]
Answer: The maximum wavelength is \(\boxed{2.48 \times 10^{-5} \text{ m}}\) (Option 3)
To find the maximum wavelength of light required to produce a hole in a p-type semiconductor, we can use the energy-wavelength relationship of photons, given by the equation:
\(E = \frac{hc}{\lambda}\)
where:
Given that the acceptor level is situated 50 meV above the valence band, first convert the energy from meV to joules:
Now, solve for \(\lambda\):
\(\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8.01 \times 10^{-21}}\)
Simplify the expression:
\(\lambda = \frac{19.878 \times 10^{-26}}{8.01 \times 10^{-21}}\)
\(\lambda \approx 2.48 \times 10^{-6} \, \text{m}\)
Thus, the maximum wavelength of light required to produce a hole is \(2.48 \times 10^{-5} \, \text{m}\), which matches one of the given options.
The correct answer is: \(2.48 \times 10^{-5} \, \text{m}\)
\(\text{ The energy required to produce a hole is } 50 \, \text{meV} = 50 \times 10^{-3} \, \text{eV} = 8.0 \times 10^{-21} \, \text{J}.\\\)
\(\text{The wavelength corresponding to this energy is given by:}\)
\(E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}\)
\(\text{Substitute } h = 6.626 \times 10^{-34} \, \text{J.s}, \, c = 3 \times 10^8 \, \text{m/s}, \text{ and } E = 8.0 \times 10^{-21} \, \text{J}:\)
\[\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8.0 \times 10^{-21}} = 2.48 \times 10^{-5} \, \text{m}\]A pure silicon crystal with 5 × 1028 atoms m−3 has ni = 1.5 × 1016 m−3. It is doped with a concentration of 1 in 105 pentavalent atoms, the number density of holes (per m3) in the doped semiconductor will be: