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astra-53 [7]
3 years ago
11

What is the radius of a circle that has a circumference of 87.92 inches

Chemistry
2 answers:
Flauer [41]3 years ago
7 0
13.99
hope this helps
Free_Kalibri [48]3 years ago
5 0
Radius would be 14. you get the circumference (87.92) and divide it by pi (3.14) you get 28 which is the diameter and divide it by 2 and you get the radius 14.
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(a) Calculate the wavelength of light in vacuum that has a frequency of 5.49 ✕ 1018 Hz. 0.0546 Correct: Your answer is correct.
vfiekz [6]

Answer:

a) Wavelength of the light in vacuum = (5.46 × 10⁻¹¹) m = 0.0546 nm

b) Wavelength of the light in diamond = (2.26 × 10⁻¹¹) m = 0.0226 nm

c) Energy of one photon in vacuum = (3.638 × 10⁻¹⁵) J = (2.271 × 10⁴) eV

d) No, the energy of the photon doesn't change when it is travelling inside diamond.

Explanation:

Wavelength (λ), frequency (f) and velocity of light (v) are related as thus

v = fλ

a) v = fλ

v = velocity of light in vacuum = (3.0 × 10⁸) m/s

f = frequency of the light = (5.49 × 10¹⁸) Hz

λ = wavelength of the light = ?

λ = (v/f) = (3.0 × 10⁸) ÷ (5.49 × 10¹⁸)

= (5.46 × 10⁻¹¹) m = 0.0546 nm

b) To find the wavelength of the light in diamond, we need the refractive index of diamond. This is because light, just like all other waves, change their velocities and subsequently their wavelengths in different materials according to the refractive index of the materials.

Refractive index of diamond = 2.42 (from literature)

2.42 = (wavelength of light in vacuum) ÷ (wavelength of light in diamond)

2.42 = 0.0546 ÷ λ

λ = 0.0546 ÷ 2.42 = 0.0226 nm

c) Energy of a photon in vacuum is given as

E = hf

where E = energy in Joules = ?

h = Planck's constant = (6.626 × 10⁻³⁴) J.s

f = frequency of the light in vacuum = (5.49 × 10¹⁸) Hz

E = (6.626 × 10⁻³⁴) × (5.49 × 10¹⁸) = (3.638 × 10⁻¹⁵) J

1 eV = (1.602 × 10⁻¹⁹) J

The amount of the calculated energy in eV

= (3.638 × 10⁻¹⁵) ÷ (1.602 × 10⁻¹⁹) = (2.271 × 10⁴) eV

d) As light travels from material to material, it's velocity and wavelength changes from material to material, but the frequency of the light waves stay the same. Since the energy of the photon depends solely on this frequency, it shows that the energy of the photon stays consistent in whichever material.

Hope this Helps!!!

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3 years ago
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