What is the energy of photons emitted by the LED at a frequency of 610 THz
The Correct Answer and Explanation is:
The energy EEE of a photon can be calculated using the formula: E=h×fE = h \times fE=h×f
Where:
- EEE is the energy of the photon.
- hhh is Planck’s constant, 6.626×10−34 J⋅s6.626 \times 10^{-34} \, \text{J} \cdot \text{s}6.626×10−34J⋅s.
- fff is the frequency of the photon.
In this case, the frequency fff is given as 610 THz (terahertz), which is equivalent to 610×1012 Hz610 \times 10^{12} \, \text{Hz}610×1012Hz.
Now, plugging the values into the equation: E=(6.626×10−34 J⋅s)×(610×1012 Hz)E = (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (610 \times 10^{12} \, \text{Hz})E=(6.626×10−34J⋅s)×(610×1012Hz) E=4.04×10−19 JE = 4.04 \times 10^{-19} \, \text{J}E=4.04×10−19J
Thus, the energy of the photons emitted by the LED at a frequency of 610 THz is approximately 4.04×10−19 J4.04 \times 10^{-19} \, \text{J}4.04×10−19J.
Explanation:
The energy of a photon is directly proportional to its frequency. This relationship is described by Planck’s equation, where E=h×fE = h \times fE=h×f. The constant hhh, known as Planck’s constant, represents the fundamental scale of quantum mechanics, and it links the frequency of light to its energy.
In this example, the LED emits light with a frequency of 610 THz, which corresponds to a visible color in the orange-red part of the electromagnetic spectrum. The higher the frequency, the greater the energy of the photons emitted. Conversely, lower-frequency light (such as red light) carries less energy per photon compared to higher-frequency light (like blue or ultraviolet light).
This principle underlies the operation of LEDs and other light-emitting devices, where the color (and therefore energy) of the emitted light is determined by the frequency of the oscillating electrical current that excites the electrons in the material.
