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frutty [35]
3 years ago
6

A pair of narrow, parallel slits separated by 0.254 mm are illuminated by green light of wavelength 559 nm. An interference patt

ern is observed on a screen 1.71 m away from the plane of the slits. Calculate the distance from the central maximum to the first bright region on either side of the central maximum. Answer in mm. Answer in units of mm.
Physics
1 answer:
UkoKoshka [18]3 years ago
4 0

Answer: 3.76 mm

Explanation: The formulae that defines the position of a nth fringe from the center of a double slit interference experiment is given below as

y =R×mλ/d

Where y = distance between the nth fringe and the center

R = distance between slits and screen = 1.71 m

m = position of fringe = 1

d = distance between slits = 0.254 mm = 0.000254m

λ = wavelength of light = 559 nm = 559×10^-9m

By substituting parameters, we have that

y = 1.71×1×559×10^-9/0.000254

y = 0.00000095589 / 0.000254

y = 0.00376m = 3.76 mm

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The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

\large {\boxed {E = h \times f}}

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem!

<u>Given:</u>

Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule

<u>Unknown:</u>

fo = ?

<u>Solution:</u>

\Phi = h \times fo

3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo

fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )

\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544
  • Photoelectric Threshold Wavelength : brainly.com/question/10015690

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

8 0
2 years ago
Read 2 more answers
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