KCET · Physics · Semiconductors
The energy gap of an LED is \(2.4 \mathrm{eV}\). When the LED is switched ON, the momentum of the emitted photons is
- A \(1.28 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- B \(2.56 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- C \(1.28 \times 10^{-11} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- D \(0.64 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(1.28 \times 10^{-11} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Energy gap of LED, \(E_g=2.4 \mathrm{eV}\)
We know that
\(E_g=\frac{h c}{\lambda} \Rightarrow \frac{h}{\lambda}=\frac{E_g}{c}\)
\(\therefore\) Momentum of emitted photon
\(\begin{aligned} p & =\frac{h}{\lambda} \\ \frac{E_g}{c} & =\frac{24 \times 1.6 \times 10^{-19}}{3 \times 10^8} \\ & =1.28 \times 10^{-11} \mathrm{~kg}-\mathrm{ms}^{-1}\end{aligned}\)
We know that
\(E_g=\frac{h c}{\lambda} \Rightarrow \frac{h}{\lambda}=\frac{E_g}{c}\)
\(\therefore\) Momentum of emitted photon
\(\begin{aligned} p & =\frac{h}{\lambda} \\ \frac{E_g}{c} & =\frac{24 \times 1.6 \times 10^{-19}}{3 \times 10^8} \\ & =1.28 \times 10^{-11} \mathrm{~kg}-\mathrm{ms}^{-1}\end{aligned}\)
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