Answer
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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
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