Question

The longest wavelength that can be absorbed by silicon, which has the bandgap of \(1.12\) eV, is \(1.1\,\mu m\). If the longest wavelength that can be absorbed by another material is \(0.78\,\mu m\), then the bandgap of this material is :

• A. \(1.416\) eV
• B. \(0.886\) eV
• C. \(1.59\) eV
• D. \(3.5\) eV

Hint:

Bandgap and wavelength relation: \[ E_g(\text{eV})=\frac{1.24}{\lambda(\mu m)} \] Higher bandgap means lower wavelength absorption limit.

Answer 1

The Correct Option is C

Concept: The relationship between bandgap energy and wavelength is: :contentReference[oaicite:1]{index=1} where:
• \(E_g\) is in eV
• \(\lambda\) is in \(\mu m\) Smaller wavelength corresponds to larger bandgap energy.

Step 1: Write the given values. For silicon: \[ E_{g1}=1.12\ \text{eV} \] \[ \lambda_1 = 1.1\ \mu m \] For the unknown material: \[ \lambda_2 = 0.78\ \mu m \] We need to calculate: \[ E_{g2} \]

Step 2: Use proportional relation. Since: \[ E_g \propto \frac{1}{\lambda} \] we can write: \[ \frac{E_{g2}}{E_{g1}} = \frac{\lambda_1}{\lambda_2} \] Substitute values: \[ \frac{E_{g2}}{1.12} = \frac{1.1}{0.78} \]

Step 3: Calculate numerical value. \[ \frac{1.1}{0.78} = 1.410 \] Hence: \[ E_{g2} = 1.12 \times 1.410 \] \[ E_{g2} = 1.579\ \text{eV} \] Approximating: \[ E_{g2}\approx1.59\ \text{eV} \]

Step 4: Write final answer. Thus the bandgap energy is: \[ \boxed{1.59\ \text{eV}} \] Hence the correct option is: \[ \boxed{(C)} \]