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Electronic Devices question

2022 · 25 Jul · Shift 1 · Q68
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Electronic Devices question

2022 · 25 Jul · Shift 1 · Q68

JEE MainPhysicsElectronic DevicesNumerical+4 / −1
The energy band gap of semiconducting material to produce violet (wavelength = 4000 Ao\mathop A\limits^oAo​) LED is ‾\underline{\hspace{2cm}}​eV\mathrm{eV}eV. (Round off to the nearest integer).
Numerical answer
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Correct answer: 3

  1. Use the relation between photon energy and band gap

For an LED, the emitted photon energy is approximately equal to the semiconductor band gap:

Eg≈E=hcλE_g \approx E = \frac{hc}{\lambda}Eg​≈E=λhc​

In electron-volts, a convenient formula is:

E(eV)=12400λ(A˚)E(\text{eV}) = \frac{12400}{\lambda(\mathring{A})}E(eV)=λ(A˚)12400​

  1. Substitute the given wavelength

Given:

λ=4000 A˚\lambda = 4000\ \mathring{A}λ=4000 A˚

So,

Eg=124004000=3.1 eVE_g = \frac{12400}{4000} = 3.1\ \text{eV}Eg​=400012400​=3.1 eV

  1. Round off to the nearest integer

3.1≈33.1 \approx 33.1≈3

Hence, the required energy band gap is:

3 eV\boxed{3\ \text{eV}}3 eV​

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