In a \(p\)-type semiconductor the donor level is at \(50 \mathrm{meV}\) above the valence band. To produce…
In a \(p\)-type semiconductor the donor level is at \(50 \mathrm{meV}\) above the valence band. To produce one electron, the maximum wavelength of light photon required is (Planck's constant, \(h=6.6 \times 10^{-34} \mathrm{Js}\) and speed of light in vacuum, \(c=3 \times 10^8 \mathrm{~ms}^{-1}\) )
\(0.0248 \mu \mathrm{m}\)
\(0248 \mu \mathrm{m}\)
\(2.48 \mu \mathrm{m}\)
\(24.8 \mu \mathrm{m}\)
Solution
Given,
\(p\)-type semiconductor donor energy level,
\(E=50 \mathrm{meV}=50 \times 10^{-3} \times 1.6 \times 10^{-19} \mathrm{~V}\)
Planck's constant, \(h=6.6 \times 10^{-34} \mathrm{Js}\)
speed of light in vacuum, \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Now, for the maximum wavelength of light photon's required \((p)\).
According to the Planck's quantum theory,
\(\therefore \quad E=h v \Rightarrow E=\frac{h c}{\lambda} \quad\left[\because v=\frac{c}{\lambda}\right]\)
Putting the given values, we get
\(50 \times 10^{-3} \times 1.6 \times 10^{-19}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{\lambda}\)
\(\begin{aligned}
\lambda & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{50 \times 10^{-3} \times 1.6 \times 10^{-19}} \\
& =\frac{6.6 \times 3 \times 10^{-34} \times 10^8}{5 \times 16 \times 10^{-22}}=\frac{6.6 \times 3 \times 10^{-4}}{5 \times 16} \\
& =2.475 \times 10^{-5} \mathrm{~m} \\
\text{or } \lambda & =24.75 \times 10^{-6}=24.75 \mu \mathrm{m}
\end{aligned}\)
Hence, the maximum wavelength of light photon required is \(24.8 \mu \mathrm{m}\).