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trasher [3.6K]
3 years ago
7

What is the wavelength of light (nm) that has a frequency of 6.44 x 1013 s-1?

Physics
2 answers:
Lostsunrise [7]3 years ago
6 0

The wavelength of the wave of light having the frequency of  6.44 \times {10^{13}}\,{\text{Hz}} will be \boxed{4.658 \times {{10}^{ - 6}}\,{\text{m}}} or  \boxed{4658\,{\text{nm}}}.

Further Explanation:

Given:

The frequency of the light is  6.44 \times {10^{13}}\,{\text{Hz}}.

Concept:

The light wave is an electromagnetic wave. An electromagnetic wave is a transverse wave. All the electromagnetic waves having different frequencies and wavelength are considered to be moving at the speed of light.

The transverse waves are the waves in which the direction of vibration of the medium particle is perpendicular to the propagation of the wave.

The light rays are called as the electromagnetic waves because they have the electric field as well as the magnetic field oscillating perpendicular to each other and also perpendicular to the direction of propagation of the wave.

The relation between the wavelength of the electromagnetic waves and its frequency is:

\boxed{\lambda=\frac{c}{f}}

Here, \lambda is the wavelength, c is the speed of light and f is the frequency of the light.

The speed of light is 3 \times {10^8}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}.

Substitute the values in above expression.

\begin{aligned}\lambda&= \frac{{3 \times {{10}^8}}}{{6.44 \times {{10}^{13}}}}\,{\text{m}}\\&= 4.{\text{658}}\times {\text{1}}{{\text{0}}^{ - 6}}\,{\text{m}}\\&= 4658\,{\text{nm}}\\\end{aligned}

Thus, the wavelength of the wave of light having the frequency of 6.44\times {10^{13}}\,{\text{Hz}} will be \boxed{4.658 \times {{10}^{ - 6}}\,{\text{m}}} or \boxed{4658\,{\text{nm}}}

Learn More:

1. Which of the following statements is a consequence of the equation e = mc2 brainly.com/question/2321496

2. What is the frequency of light for which the wavelength is 7.1 × 102 nm brainly.com/question/9559140

3. A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106hz and 1hz=1s−1 brainly.com/question/9527365

Answer Details:

Grade: High School

Subject: Physics

Chapter: Electromagnetic waves

Keywords:  Wavelength, frequency, electromagnetic waves, transverse waves, light wave, electric and magnetic field, oscillating perpendicular, 6.44 x 10^13 Hz.

lesantik [10]3 years ago
4 0
4660

Hope this helped!
STSN
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insens350 [35]

Answer:

774.8 secs

Explanation:

distance(d)= speed(v)* time(t)

calculate speed:

refractive index = speed of light (c)/ speed of light in medium (v)

1.56 = 3*10^8*v

v=192307692.3 m/s

d = v *t

t = d/v

on substituting values:

t = 774.8 secs

4 0
2 years ago
Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
3 years ago
Which of the following has potential but not kinetic energy?
joja [24]

Answer:

<h2>D.)</h2>

Explanation:

Potential energy is when a object is not in motion and the ball is sitting on a shelf not being thrown around or rolling. kinetic energy is when a obeject is in motion and moving.

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Question 5
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Answer:

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3 years ago
Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

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3 years ago
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