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Standard XII

Physics

Resonance in RLC Circuit

The band gap ...

Question

The band gap in $eV$ of a semiconductor material required to construct an LED that emits peak power at the wavelength of $620 nm$ is

$(Plank constant h = 4.13567 \times 10^{- 15} eVs$ $and speed of light c = 2.998 \times 10^{8} m/s)$
2

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Solution

The correct option is A 2
$E_{g} = \frac{h c}{λ} = \frac{4.13567 \times 10^{- 15} \times 2.998 \times 10^{8}}{620 \times 10^{- 9}}$
$= 2 eV$

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The band gap in $eV$ of a semiconductor material required to construct an LED that emits peak power at the wavelength of $620 nm$ is

$(Plank constant h = 4.13567 \times 10^{- 15} eVs$ $and speed of light c = 2.998 \times 10^{8} m/s)$
2

The correct option is A 2
$E_{g} = \frac{h c}{λ} = \frac{4.13567 \times 10^{- 15} \times 2.998 \times 10^{8}}{620 \times 10^{- 9}}$
$= 2 eV$

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The band gap in eV of a semiconductor material required to construct an LED that emits peak power at the wavelength of 620 nm is (Plank constant h=4.13567×10−15 eVs and speed of light c=2.998×108 m/s)2

The correct option is A 2Eg=hcλ=4.13567×10−15×2.998×108620×10−9 =2 eV

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EXAMS

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FREE TEXTBOOK SOLUTIONS

STATE BOARDS

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The band gap in $eV$ of a semiconductor material required to construct an LED that emits peak power at the wavelength of $620 nm$ is

$(Plank constant h = 4.13567 \times 10^{- 15} eVs$ $and speed of light c = 2.998 \times 10^{8} m/s)$
2

The correct option is A 2
$E_{g} = \frac{h c}{λ} = \frac{4.13567 \times 10^{- 15} \times 2.998 \times 10^{8}}{620 \times 10^{- 9}}$
$= 2 eV$