Question

In a p-type semiconductor, the acceptor level is situated 50 meV above the valence band. The maximum wavelength of light required to produce a hole will be:

• A. $24.8 \times 10^{-5} \, \text{m} \\$
• B. $0.248 \times 10^{-5} \, \text{m} \\$
• C. $2.48 \times 10^{-5} \, \text{m} \\$
• D. $248 \times 10^{-5} \, \text{m}$

Hint:

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)

Answer 1

The Correct Option is C

To determine the maximum wavelength of light capable of creating a hole in a p-type semiconductor, we employ the photon energy-wavelength relationship:

\(E = \frac{hc}{\lambda}\)

where:

• \(E\) represents the required energy,
• \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}\)),
• \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
• \(\lambda\) is the wavelength.

Given that the acceptor level is located 50 meV above the valence band, convert this energy from meV to joules:

• \(50 \, \text{meV} = 50 \times 10^{-3} \, \text{eV} = 50 \times 10^{-3} \times 1.602 \times 10^{-19} \, \text{J}\)
• \( = 8.01 \times 10^{-21} \, \text{J}\)

Rearrange the formula to solve for \(\lambda\):

\(\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8.01 \times 10^{-21}}\)

Perform the calculation:

\(\lambda = \frac{19.878 \times 10^{-26}}{8.01 \times 10^{-21}}\)

\(\lambda \approx 2.48 \times 10^{-6} \, \text{m}\)

Therefore, the maximum wavelength of light necessary for hole generation is approximately \(2.48 \times 10^{-5} \, \text{m}\), aligning with one of the provided options.

The correct answer is: \(2.48 \times 10^{-5} \, \text{m}\)