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Step-by-Step Solution
Step 1: Identify the Relevant Physical Quantity
The problem states that a semiconductor's conductivity increases when it is exposed to electromagnetic radiation of wavelength shorter than 2480 nm. This wavelength threshold corresponds to the photon energy that is equal to the band gap energy of the semiconductor.
Step 2: Write Down the Photon Energy Formula
The energy of a photon, in joules, is given by:
$E = \frac{hc}{\lambda}$
where:
$h$ is Planck’s constant ($6.63 \times 10^{-34}\, \text{J}\cdot \text{s}$),
$c$ is the speed of light in vacuum ($3.0 \times 10^{8}\, \text{m/s}$),
$\lambda$ is the wavelength of the radiation in meters.
Step 3: Convert the Wavelength to Meters
Given $\lambda = 2480\,\text{nm}$, convert it into meters:
$2480\,\text{nm} = 2480 \times 10^{-9}\,\text{m}$.
Step 4: Substitute Values to Get Energy in Joules
$E = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{2480 \times 10^{-9}} \,\text{J}$
Step 5: Convert the Energy from Joules to Electron Volts
Since $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$, we divide the energy in joules by $1.6 \times 10^{-19}\,\text{J/eV}$ to find $E$ in eV:
$E_{\text{in eV}} = \left(\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{2480 \times 10^{-9}}\right) \times \frac{1}{1.6 \times 10^{-19}}$
Step 6: Simplify the Expression
Evaluating this expression gives approximately:
$E \approx 0.5\, \text{eV}.$
Step 7: State the Final Answer
The band gap for the semiconductor is $0.5\,\text{eV}$.
