NEETAtomic StructureHardBloom L3
Question
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480 \text{ nm}$ is incident on it. The band gap (in eV) for the semiconductor is:
Options
A.$0.9 \text{ eV}$
B.$0.7 \text{ eV}$
C.$0.5 \text{ eV}$
D.$1.1 \text{ eV}$
Solution
{"given":"Maximum wavelength causing increased conductivity: $\\lambda_{\\max} = 2480 \\text{ nm} = 24800 \\text{ Å}$. The band gap energy equals the minimum photon energy required to excite electrons across the gap, i.e., the energy corresponding to $\\lambda_{\\max}$.","key_observation":"The photon energy formula $E = \\dfrac{12400}{\\lambda(\\text{Å})} \\text{ eV}$ gives the minimum energy (band gap) when the maximum wavelength is used.","option_analysis":[{"label":"(A)","text":"0.9 eV","verdict":"incorrect","explanation":"This would correspond to $\\lambda = 12400/0.9 \\approx 13778$ Å $\\approx 1378$ nm, not 2480 nm."},{"label":"(B)","text":"0.7 eV","verdict":"incorrect","explanation":"This would correspond to $\\lambda = 12400/0.7 \\approx 17714$ Å $\\approx 1771$ nm, not 2480 nm."},{"label":"(C)","text":"0.5 eV","verdict":"correct","explanation":"$$E = \\frac{12400}{24800} = 0.5 \\text{ eV}$$, which correctly matches $\\lambda_{\\max} = 2480 \\text{ nm} = 24800 \\text{ Å}$."},{"label":"(D)","text":"1.1 eV","verdict":"incorrect","explanation":"This would correspond to $\\lambda = 12400/1.1 \\approx 11273$ Å $\\approx 1127$ nm, not 2480 nm."}],"answer":"(C)","formula_steps":[]}
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