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AfilCa [17]
2 years ago
7

If a photon has a frequency of 5.20 × 1014 hertz, what is the energy of the photon? given: planck's constant is 6.63 × 10-34 jou

le·seconds. a. 3.45 × 10-19 joules b. 7.91 × 10-16 joules c. 4.32 × 10-14 joules d. 3.43 × 10-14 joules
Physics
1 answer:
Inessa05 [86]2 years ago
4 0

If a photon has a frequency of 5.20 × 1014 hertz, then the energy of the photon will be 3.45 x 10^-19 Joules.

<h3>What is photon energy?</h3>

The energy carried by a single photon is referred to as photon energy. The amount of energy is related to the electromagnetic frequency of the photon and consequently <u>inversely proportional t</u>o the wavelength. The higher the frequency of a photon, the greater its energy. The longer the wavelength of a photon, the lower its energy.

<h3>Given: </h3>

frequency = 5.20 × 10^(14) hertz

plank's constant = 6.63 × 10^-34 joule·seconds

<h3>Formula used:</h3>

E = hv

( where E = energy of the photon)

<h3>Solution: </h3>

E = 6.63 x 10 ^(-34) x 5.20 x 10^(14) = 34.476 x 10^ -19 Joules

To learn more about photons :

brainly.com/question/20912241

#SPJ4

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To be able to answer this item, we are to calculate the power that the machine could deliver from hp to kW. 

      (45 hp)(746 W/1 hp) = 33570 W

Power is the amount of energy delivered at a certain period. 

             t = (6.20 x 10^2 J)/ (33570 kJ/s)

             t = 0.01845 s
7 0
3 years ago
Why are paperclips attracted to magnets? Group of answer choices They contain silver They contain aluminum They contain tin They
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Answer:

They contain iron.

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5 0
3 years ago
What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?
ivann1987 [24]

Answer:

2.46\cdot 10^5 J

Explanation:

The energy of a single photon is given by:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength

For the photon in this problem,

\lambda=486 nm=4.86\cdot 10^{-7}m

So, its energy is

E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J

One mole of photons contains a number of photons equal to Avogadro number:

N_A = 6.022\cdot 10^{23}

So, the total energy of one mole of photons is

E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J

7 0
4 years ago
Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio
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Answer:

B ) Ascend using my buddy alternative air source / make an emergency Ascent

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5 0
3 years ago
Read 2 more answers
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
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Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
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