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Physics · Semiconductor and Electronic Devices

JEE Main 2024 — 6 April, Shift 2 — Question 36

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

  1. Option A:

    407 nm

  2. Option B:

    414 nm

  3. Option C:

    207 nm

    Correct
  4. Option D:

    103.5 nm

Answer: C

Step-by-step solution

\quad Energy =hcλ=\frac{\mathrm{hc}}{\lambda}; E=1240λ( nm)eV\mathrm{E}=\frac{1240}{\lambda(\mathrm{~nm})} \mathrm{eV}

6=1240λ( nm)6=\frac{1240}{\lambda(\mathrm{~nm})}

λ=12406=207 nm\lambda=\frac{1240}{6}=207 \mathrm{~nm}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 2
Subject
Physics
Chapter
Semiconductor and Electronic Devices
Topic
p-n Diode and its Applications
The acceptor level of a p-type semiconductor is 6 eV . The maximum… | JEE Main 2024 PYQ with Solution · DhiX AI