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3 years ago

What does a model of a light wave tell us about brightness and color?

Physics

1 answer:

tino4ka555 [31]3 years ago

3 0

Answer:

A wave model of light is useful for explaining brightness,color, and the frequency-dependent bending of light at a surface between media. However, because light can travel through space, it cannot be a matter wave, like sound or water waves.

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"Compared to infrared radiation, does ultraviolet radiation have longer or shorter wavelengths? Does ultraviolet radiation have

balu736 [363]

Answer:Ultraviolet radiation has shorter wavelengths and higher energy than infrared radiation.

Explanation: Electromagnetic radiation radiations which have both electrical and magnetic properties,they can be transmitted through space or through a medium.

It includes Gamma radiation, infra-red, visible light, Ultraviolet radiation etc they occur with different wavelength, the lower the wavelength the higher the Energy dissipated per photon. According to their order of decreasing wavelength and increased energy they are classified as follows.

RADIO WAVE, MICRO WAVE, INFRA-RED, VISIBLE LIGHT, ULTRAVIOLET RAY, X-RAY, GAMMA RAYS.

5 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

A boy notices that 45 complete waves travel past him in ninety seconds. What is the

san4es73 [151]

Answer:

0.5 Hz

Hope you find this helpful. Please mark me as brainliest!

5 0

3 years ago

Which is formed by positive ions from a base and negative ions from an acid?

Lostsunrise [7]

The answer is: Salt! :)

Have a great day!

6 0

3 years ago

Read 2 more answers

A bus is traveling with a uniform acceleration of 2.75 meters/second2. If the initial velocity of the bus is 16.5 meters/second,

Kaylis [27]

<span>The velocity will be 41.25 m/s2 after 9 seconds. To find velocity after a specific time period, multiply the acceleration (2.75) times the number of seconds (9) to receive 24.75 m/s, then add that to the initial velocity of 16.5 m/s. 24.75 + 16.5 = 41.25 m/s2.</span>

8 0

3 years ago