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3 years ago

¿Cuál es el rango de frecuencias comprendido entre las longitudes de onda de 220 nm, 350 nm,

Physics

1 answer:

olchik [2.2K]3 years ago

3 0

Answer:

f1 = 1.36*10^15 Hz

f2 = 8.57*10^14 Hz

f3 = 3.75*10^14 Hz

f4 = 8.57*10^13 Hz

Explanation:

To know which are the frequency ranges of the wavelengths you use the following formula:

$c=\lambda f\\\\f=\frac{c}{\lambda}$

c: speed of light = 3*10^8m/s

λ: wavelength = 220nm, 350nm, 800nm, 3500nm

f: frequency of the wave

You calculate f for each wavelength, furthermore, you add the kind of electromagnetic wave:

λ1 = 220nm = 220*10^-9 m

$f=\frac{3*10^8m/s}{220*10^{-9}m}=1.36*10^{15}Hz$ ULTRAVIOLET

λ2 = 350nm = 350*10^-9 m

$f=\frac{3*10^8m/s}{350*10^{-9}m}=8.57*10^{14}Hz$ VISIBLE LIGHT

λ3 = 800nm = 350*10^-9 m

$f=\frac{3*10^8m/s}{800*10^{-9}m}=3.75*10^{14}Hz$ INFRARED

λ4 = 3500nm = 350*10^-9 m

$f=\frac{3*10^8m/s}{3500*10^{-9}m}=8.57*10^{13}Hz$ MICROWAVE - INFRARED

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An LED with total power P tot = 960 mW emits UV light of wavelength 360 nm. Assuming the LED is 55% efficient and acts as an iso

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Answer:

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Explanation:

We are given that

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Distance,r=2.5 cm= $2.5\times 10^{-2} m$

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$I=\frac{0.528}{4\pi(2.5\times 10^{-2})^2}=67.26W/m^2$

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$E^2_0=\frac{2I}{c\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}$

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$E_0=\sqrt{\frac{2\times 67.26}{3\times 10^8\times 8.85\times 10^{-12}}}$

$E_0=225.09N/C$

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