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konstantin123 [22]
2 years ago
9

(i) Calculate the energy, in electron volts, of a photon whose frequency is (a) 620 THz}, (b) 3.10GHz , and(c) 46.0 MHz

Physics
1 answer:
DedPeter [7]2 years ago
6 0

The energy in electron volts of the photons that has the following frequencies is as follows:

  1. 620 THz = 2.564eV
  2. 3.10GHz = 1.28 × 10-⁵eV
  3. 46.0 MHz = 1.902 × 10-⁷eV

<h3>How to calculate energy?</h3>

The energy of a photon can be calculated using the following formula:

E = hf

Where;

  • E = energy
  • h = Planck's constant (6.626 × 10-³⁴ J/s)
  • f = frequency

First, we convert the frequencies to hertz as follows;

  • 620THz = 6.2 × 10¹⁴Hz
  • 3.10GHz = 3.1 × 10⁹Hz
  • 46.0MHz = 4.6 × 10⁷Hz

  1. E = 6.626 × 10-³⁴ × 6.2 × 10¹⁴ = 2.564eV
  2. E = 6.626 × 10-³⁴ × 3.1 × 10⁹ = 1.28 × 10-⁵eV
  3. E = 6.626 × 10-³⁴ × 4.6 × 10⁷ = 1.902 × 10-⁷eV

Learn more about energy of a photon at: brainly.com/question/2393994

#SPJ1

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An overhead projector lens is
Inessa [10]

Answer: 34.9 cm

Explanation:

You are given the following parameters;

Object distance U = 32 cm

Magnification M = - 12.0

According to formula for magnification;

M = V/U

Where V = image distance.

Substitute V and M into the formula

-12 = V/32

Cross multiply

V = -12 × 32

V = - 384

You can use the formula

1/f = 1/V + 1/U

Where f = focal length

Substitute V and U into the formula

1/f = - 1/384 + 1/32

Find the lowest common factor of the denominator at Left hand side

1/f = ( -1 + 12 ) / 384

1/f = 11/384

Reciprocate both sides

F = 384/11

F = 34.9 cm

He should therefore use the focal length of 34.9 cm

6 0
3 years ago
What is the frequency of radiation whose wavelength is 11.5 a0 ?
irakobra [83]

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

4 0
3 years ago
Why are multiple images seen when two plane mirrors are placed at an angle​
Whitepunk [10]

Answer: Can I get a picture???

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A 0.150 kg mass is attached to a spring with k = 18.9 N/m. At the equilibrium position, it moves 2.39 m/s. How much mechanical e
Nastasia [14]

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Explanation: got it right on Acellus

5 0
3 years ago
Which of the following would be used in luminosity calculations?
Daniel [21]

Answer:

1) joule

2) kgm^{2}/s^{2}

3) 10\%

Explanation:

1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second (\frac{J}{s}), which is equal to Watts (W).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work W is expressed as force  F multiplied by the distane  d :

W=F.d

Where force has units of  kgm/s^{2} and distance units of m.

If we input the units we will have:

W=(kgm/s^{2})(m)

W=kgm^{2}/s^{2}  This is 1Joule (1 J) in the SI system, which is also equal to 1 Nm

3) The formula to calculate the percent error is:

\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%

Where:

V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2} is the experimental value

V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2} is the accepted value

\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%

\% error=10.04\% \approx 10\% This is the percent error

8 0
4 years ago
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