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Agata [3.3K]
3 years ago
7

What is the photon energy (eV) of indigo light​

Physics
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

When dealing with "particles" such as photons or electrons, a commonly used unit of energy is the electron-volt (eV) rather than the joule (J). An electron volt is the energy required to raise an electron through 1 volt, thus a photon with an energy of 1 eV = 1.602 × 10-19 J.

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Mrs. Lynn's art class is making pencil holders out of soup cans. Each student is first covering a can with felt. What will the c
DochEvi [55]

Answer:

25.13 cm

Explanation:

Mrs. Lynn's art class is making pencil holders out of soup cans

These soup cans are definitely in cylindrical forms.

Each student is first covering a can with felt.

A felt is a type of material that is used for covering other objects. Here a felt is used to cover soup cans.

So if the diameter of the bottom of the can is 8 cm (i.e d = 8cm)

What will the circumference of the piece of felt need to be in order to cover the bottom of the can?

The felt will also take the circumference of a cylinder; and which is given by the expression

C = 2 πr

since diameter (d) = 8 cm

radius (r) = \frac{d}{2}

r = \frac{8}{2}

r = 4 cm

C = 2 × π × 4 cm

C = 2 × 3.142 × 4 cm

C = 25.13 cm

∴ the circumference of the piece of felt need to be 25.13 cm in order to cover the bottom of the can.

6 0
3 years ago
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou
SCORPION-xisa [38]

Answer:

imma try and get it wrong

3 0
2 years ago
what is a point of view of an object used to determine another obejects motion i nedd help asap plsss​
Igoryamba
A believe that’s called a reference point.
9 0
3 years ago
Alexis is studying how lenses work. She looks through a
Kryger [21]
It would be the 3rd one
6 0
3 years ago
A LED light source contains a 0.5-Watts GaAs (Eg =1.43 eV) LED. Assuming that 0.12% of the electric energy is converted to emiss
Ray Of Light [21]

Answer:

Explanation:

energy emitted by  source per second  = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ  ;  h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵  x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ  , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6  x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10  x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵   ;   decibel scale is logarithmic , I is energy of sound having dB = 50 and  I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

= 10 x 10⁻⁸ W

power required

= 10⁻⁸ + 10 x 10⁻⁸ W

= 11  x 10⁻⁸ W.

5 0
3 years ago
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