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Readme [11.4K]
2 years ago
5

What is the energy of a photon with a frequency of 3. 6 Ă— 1015 Hz? Planck’s constant is 6. 63 Ă— 10â€""34 J•s. 1. 8 Ă— 10â€

""49 J 2. 4 Ă— 10â€""19 J 1. 8 Ă— 10â€""18 J 2. 4 Ă— 10â€""18 J.
Physics
1 answer:
erastovalidia [21]2 years ago
8 0

Every photon has some energy within it and that energy is called photon energy.

The energy of photon is 2.38\times10^{-18}\;\rm J.

Given that, frequency of the photon is 3.6\times10^{15}\;\rm Hz and Planck's constant is 6.626\times {{10}^{-34}}.

So the energy of the photon can be calculated by the given formula.

E=h\nu

Where h is plank's constant and \nu is the frequency of the photon.

By substituting the values in the above formula, the photon energy is,

E=6.626\times10^{-34}\times3.6\times10^{15}\;\rm J

E=2.38\times10^{-18}\;\rm J

The energy of photon is 2.38\times10^{-18}\;\rm J.

For more details about the energy of photon, follow the link given below.

brainly.com/question/15870724.

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Place the items in order from the largest wavelength to the shortest wavelength.
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From largest to shortest wavelength:

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Explanation:

Electromagnetic waves are oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

Electromagnetic waves are the only type of waves able to travel in a vacuum, and in a vacuum they always at the same speed, the speed of light,  equal to:

c=3.0\cdot 10^8 m/s

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From slongest to shortest wavelength, they are ranked as follows:

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#LearnwithBrainly

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2 years ago
(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104
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Answer:

The answer is:

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7 0
2 years ago
If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t
Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to  Biot-Savart law, A magnetic field generated by a current  carrying wire at a distance is represented as

B=μ₀I/ 2πr

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 μ₀ =permeability of free space  4π × 10−7 H/m

I = current intensity

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6.50 X 10^-3 =  μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50  X 10 ^-3 X 2π  X  X 12.8 X 10^-2/  4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

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