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Andreas93 [3]
3 years ago
15

The wavelength of some yellow light is 580.0 nm. What is the frequency of this yellow light?

Chemistry
2 answers:
Ronch [10]3 years ago
7 0

The frequency of yellow light of wavelength 580 nm is  \boxed{{0.05172\times10^{18}}\text{Hz}}

Further Explanation:

Wavelength is the characteristic property of a wave. The term wavelength is defined as the distance between two successive crests or troughs in a wave. It is represented by \lambda and its SI unit is meter (m).

The term frequency is defined as the number of times n event occurs in unit time. It is generally applied to waves including light, sound and radio waves. It is denoted by {\text{v }} and its SI unit is Hertz (Hz).

The conversion factor to convert nm to m is,

1\;{\text{nm}}={10^{ - 9}}\;{\text{m}}

The wavelength and frequency of light are related to each other by the following expression,

{\text{c}} = {v\lambda}                                   ......(1)

Here,

C is the speed of light.

v is the frequency of light.

\lambda is the wavelength of light.

Rearrange equation (1) to calculate the frequency of light.

v=\dfrac{c}{\lambda}                      ......(2)

The wavelength of yellow light is 580 nm.

Firstly, wavelength of yellow light is to be converted from nm to m as follows:

\begin{aligned}{\text{Wavelength of yellow light }}\left({\text{m}}\right)&=\left( {{\text{580 nm}}}\right)\left({\frac{{{{10}^{ - 9}}\;{\text{m}}}}{{1\;{\text{nm}}}}}\right)\\&=58\times {10^{ - 10}}\;{\text{m}}\\\end{aligned}

The speed of yellow light is 3 \times {10^8}\;{\text{m/s}}.

The wavelength of yellow light is  58\times {10^{ - 10}}\;{\text{m}}.

Substitute these values in equation (2)

\begin{aligned}{v }}&=\frac{{{\text{3}}\times{\text{1}}{{\text{0}}^8}\;{\text{m/s}}}}{{58 \times{{10}^{ - 10}}\;{\text{m}}}}\\&=0.05172\times{\text{1}}{{\text{0}}^{18}}\;{\text{Hz}}\\\end{aligned}

So the frequency of yellow light is {\mathbf{0}}{\mathbf{.05172 \times 1}}{{\mathbf{0}}^{{\mathbf{18}}}}\;{\mathbf{Hz}}.

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: frequency, wavelength, yellow light, 580 nm, crests, troughs, speed of light, wavelength of light, frequency of light, conversion factor, Hz, m.

Tema [17]3 years ago
4 0

Answer : The frequency of this yellow light is, 5.172\times 10^{14}s^{-1}

Explanation : Given,

Wavelength of yellow light = 580nm=580\times 10^{-9}m

conversion used : 1nm=1\times 10^{-9}m

Formula used :

\nu=\frac{c}{\lambda}

where,

\nu = frequency of light

\lambda = wavelength of light

c = speed of light = 3\times 10^8m/s

Now put all the given values in this formula, we get:

\nu=\frac{3\times 10^8m/s}{580\times 10^{-9}m}=5.172\times 10^{14}s^{-1}

Therefore, frequency of this yellow light is, 5.172\times 10^{14}s^{-1}

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Please help me! I overthink these types of questions and I can't figure this out or remember.
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Answer:

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43 =  λ= 1.1 × 10⁻¹² m.

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Explanation:

Q 42 = What is the frequency of ultraviolet light with wavelength 2.94 × 10⁻⁸ m.

Given data:

wavelength of radiation = 2.94 × 10⁻⁸ m.

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Frequency = ?

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Formula:

speed or velocity = wavelength × frequency

c = λ × f

f = c/ λ

f = 3 × 10⁸ m/s /2.94 × 10⁻⁸ m.

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43 = what is the wavelength of gamma ray with the frequency 2.73 × 10²⁰ Hz.

Given data:

wavelength of radiation = ?

Speed of light = 3 × 10⁸ m/s

Frequency = 2.73 × 10²⁰ Hz.

Solution:

Formula:

speed or velocity = wavelength × frequency

c = λ × f

λ = c / f

λ = 3 × 10⁸ m/s /2.73 × 10²⁰ s⁻¹

λ= 1.1 × 10⁻¹² m.

So the wavelength of gamma radiation is 1.1 × 10⁻¹² m.

44= Consider an element Z that has two natural occuring elements with the following percent abundances: the isotope with the mass number of 19.0 is  56.8% abundant and the isotope with mass number of 21 is 43.2% abundant. what is average atomic mass of elements Z.

Given data:

Abundance of 1st isotope = 56.8%

Abundance of second isotope = 43.2%

Atomic mass of 1st isotope = 19 amu

Atomic mass of second isotope = 21 amu

Average atomic mass = ?

Solution:

Average atomic mass of carbon = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass of Z = (19.0 × 56.8) + (21  ×43.2) /100

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Solution:

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Q = 46= What is electronic configuration for phosphorus?

Electronic configuration:

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Abbreviated electronic configuration:

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Properties and uses;

Phosphorus is the member of nitrogen family.

It is multivalent nonmetal.

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It have five valance electrons.

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Its melting point is 44.1 °C

Its boiling point is 280 °C.

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