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So, the frequency of that light approximately [tex] \sf{\bold{5.93 \times 10^{14} \: Hz}} [/tex]
Hi ! Here I will help you to discuss the relationship between frequency and wavelength, with the velocity constant of electromagnetic waves in a vacuum. We all know that regardless of the type of electromagnetic wave, it will have the same velocity as the speed of light (light is part of electromagnetic wave too), which is 300,000 km/s or [tex] \sf{3 \times 10^8} [/tex] m/s. As a result of this constant property, the shorter the wavelength, the greater the value of the electromagnetic wave frequency. This relationship can also be expressed in this equation:
[tex] \boxed{\sf{\bold{c = \lambda \times f}}} [/tex]
With the following condition :
We know that :
What was asked :
Step by step :
[tex] \sf{c = \lambda \times f} [/tex]
[tex] \sf{3 \times 10^8 = 5.06 \times 10^{-7} \times f} [/tex]
[tex] \sf{f = \frac{3 \times 10^8}{5.06 \times 10^{-7}}} [/tex]
[tex] \sf{f \approx 0.593 \times 10^{8 - (-7)}} [/tex]
[tex] \sf{f \approx 0.593 \times 10^{15}} [/tex]
[tex] \boxed{\sf{f \approx 5.93 \times 10^{14} \: Hz}} [/tex]
So, the frequency of that light approximately [tex] \sf{\bold{5.93 \times 10^{14} \: Hz}} [/tex]