
An LED (Light Emitting Diode) is constructed from a p-n based junction on a certain $Ga - As - p$ semi conducting material whose energy gap is $1.9eV$. What is the wavelength of the emitted light?
Answer
488.1k+ views
Hint: In a LED, an electron jumps from high potential to low potential due to applied voltage. When electrons jump from high potential to low potential, light is emitted in the form of photons and the energy of each photon is equal to the energy gap. Also, energy of photon is given by $E = hv$, where $h$ is Planck's constant and $v$ is frequency of light wave.
Complete step-by-step solution:
From an LED, light is emitted in the form of light-waves (photons). Energy of each photon is equal to the energy gap of the LED.
Given, energy gap $E = 1.9eV$.
Let $\lambda $ be the wavelength of light.
Energy of photons is given by $E = hv$, where $h$ is Planck's constant and $v$ is frequency of light wave.
Also, $E = \dfrac{{hc}}{\lambda }$, where $c$ is the speed of light.
Then, $\lambda = \dfrac{{hc}}{E}$
We know that $hc = 1240eV$ and given $E = 1.9ev$, and we get wavelength in a nanometer.
After putting values in above equation, we get
$\lambda = \dfrac{{1240}}{{1.9}} \simeq 650nm$.
Then the wavelength of emitted light is $650nm$.
Hence, the correct answer is option A.
Note:- An LED produces light when a voltage is applied through the ends of the led. Colour of emitted light is dependent on the energy gap of led. As we know, different colours have different frequency and frequency depends upon energy gap then we can change colour of led by changing energy gap of led. Working of solar panel is reverse of working of led, when light waves fall on solar panel above a threshold frequency it generates voltage and current
Complete step-by-step solution:
From an LED, light is emitted in the form of light-waves (photons). Energy of each photon is equal to the energy gap of the LED.
Given, energy gap $E = 1.9eV$.
Let $\lambda $ be the wavelength of light.
Energy of photons is given by $E = hv$, where $h$ is Planck's constant and $v$ is frequency of light wave.
Also, $E = \dfrac{{hc}}{\lambda }$, where $c$ is the speed of light.
Then, $\lambda = \dfrac{{hc}}{E}$
We know that $hc = 1240eV$ and given $E = 1.9ev$, and we get wavelength in a nanometer.
After putting values in above equation, we get
$\lambda = \dfrac{{1240}}{{1.9}} \simeq 650nm$.
Then the wavelength of emitted light is $650nm$.
Hence, the correct answer is option A.
Note:- An LED produces light when a voltage is applied through the ends of the led. Colour of emitted light is dependent on the energy gap of led. As we know, different colours have different frequency and frequency depends upon energy gap then we can change colour of led by changing energy gap of led. Working of solar panel is reverse of working of led, when light waves fall on solar panel above a threshold frequency it generates voltage and current
Recently Updated Pages
Master Class 12 Social Science: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Class 10 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
The gas that burns in oxygen with a green flame is class 12 chemistry CBSE

The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE
