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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Semiconductors

JEE Main 2024 · 6 April · Evening Shift · Physics · Semiconductors

The acceptor level of a p-type semiconductor is 6 eV . The maximum...

The acceptor level of a p-type semiconductor is 6eV6 \mathrm{eV}. The maximum wavelength of light which can create a hole would be : Given hc=1242eVnm\mathrm{hc}=1242 \mathrm{eV} \mathrm{nm}.

Options

  1. A414 nm414 \mathrm{~nm}
  2. B103.5 nm103.5 \mathrm{~nm}
  3. C207 nm207 \mathrm{~nm}
  4. D407 nm407 \mathrm{~nm}

Correct answer

C. 207 nm207 \mathrm{~nm}

Step-by-step solution

 Energy =hcλE=1240λ(nm)eV6=1240λ(nm)λ=12406=207nm\begin{aligned} & \text{ Energy }=\frac{\mathrm{hc}}{\lambda} \\ & \mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV} \\ & 6=\frac{1240}{\lambda(\mathrm{nm})} \\ & \lambda=\frac{1240}{6}=207 \mathrm{ nm}\end{aligned}
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