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3 years ago

What does the wavelength of visible light waves correspond to?

Physics

2 answers:

marissa [1.9K]3 years ago

7 0

Explanation:

Visible light is a very part of electromagnetic spectrum. We can see the visible region of the electromagnetic spectrum.

The frequency and the wavelength has an inverse relationship.

For example, the red light has large wavelength and low frequency. Violet has a short wavelength and higher frequency.

The frequency range of visible region is from 400 nm to 700 nm.

Lesechka [4]3 years ago

7 0

The wavelength of the visible range of the electromagnetic spectrum lies between $\boxed{400\,{\text{nm}} - 700\,{\text{nm}}}$ .

Further Explanation:

The Electromagnetic spectrum is the complete range of the electromagnetic radiation which expresses all the radiation starting from the Gamma rays to the radio waves. The electromagnetic spectrum is the detailed spectrum that explains all the electromagnetic radiations.

The electromagnetic radiations are the transverse waves which do not require any medium for their propagation. The electromagnetic waves can easily travel through space. All the electromagnetic radiations travel at the speed of light.

The different radiations in the electromagnetic spectrum are as follows:

Gamma rays

X-rays

Ultraviolet rays

Infrared rays

Micro waves

Radio waves

The wavelength and the frequency ranges for the different electromagnetic radiations is as shown in figure attached.

Therefore, the wavelength of the visible range of the light in the electromagnetic spectrum is $\boxed{400\,{\text{nm}} - 700\,{\text{nm}}}$ .

Learn More:

1. What is the threshold frequency ν0 of cesium brainly.com/question/6953278

2. Calculate the wavelength of an electron (m = 9.11 × 10-28 g) moving at 3.66 × 106 m/s brainly.com/question/1979815

3. What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays brainly.com/question/9059731

Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetic Waves

Keywords:

Electromagnetic spectrum, visible light, EM waves, wavelength, 700nm, 400 nm, 7000 A, frequency, gamma rays, transverse waves.

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Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

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m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

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A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

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K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

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K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

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