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galina1969 [7]
2 years ago
11

What is the threshold frequency ν0 of cesium? note that 1 ev (electron volt)=1.60×10−19 j. express your answer numerically in he

rtz?

Physics
2 answers:
andreyandreev [35.5K]2 years ago
8 0

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

Katena32 [7]2 years ago
6 0

The threshold frequency of the cesium is \fbox{\begin\\5.06 \times {10^{14}}\,{\text{Hz}}\end{minispace}}.

Further Explanation:

The threshold energy is the amount of energy required by an electron so that it is ejected from the surface of the material.

Concept:

The photoelectric effect is the process in which emission of the electron from the surface of material when they acquire sufficient amount of energy from the photon (i.e. the packets of energy) to release from the surface of the material.

The threshold energy of the electron can be expressed as:

\fbox{\begin\\E = {E_k} + \phi\end{minispace}}

Here, E is the amount of energy supplied to the surface by a photon, {E_k} is the amount of kinetic energy of electron after emission and \phi is the work function of the material.

The work function of a surface is the minimum amount of energy to be incident on the surface of a metal such that it can emit the electron.

The work function of material can be expressed as:

\phi=h{\nu _0}

Here, {\nu _0} is the threshold frequency of the electron.

The work function of the cesium is 2.1\,{\text{eV}}.

Convert the work function in eV to Joules.

\begin{aligned}\phi&=2.1{\mkern1mu} {\text{eV}}\left({1.6 \times{{10}^{ - 19}}}\right)\,{\text{J}}\\&=3.36\times10^{-19}\text{ J}\\\end{aligned}

Substitute 3.36\times{10^{-19}}\,{\text{J}} for \phi and 6.63\times{10^{-34}\text{ kg.m}^2/\text{s} for h in expression for work function.

\begin{aligned}3.36\times{10^{-19}}&=\left({6.63\times{{10}^{-34}}}\right){\nu_0}\\{\nu_0}&=\frac{{3.36\times{{10}^{-19}}}}{{6.63\times{{10}^{-34}}}}\\&=5.06\times{10^{14}}\,{\text{Hz}}\\\end{aligned}

Thus, the threshold frequency of cesium is 5.06 \times {10^{14}}\,{\text{Hz}}.

Learn More:

1.  The average translational kinetic energy of molecules brainly.com/question/9078768

2.  Collision of car with a wall brainly.com/question/9484203

3. Conservation of energy brainly.com/question/3943029

Answer Details:

Grade: College

Subject: Physics

Chapter: Photoelectric effect

Keywords:

Cesium, work function, threshold frequency, photoelectric effect, photon, minimum energy, surface, frequency, eV, Joule.

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(d) 0.07 m/s

Explanation:

Given Data

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m_{1}v_{1}=m_{2}v_{2}\\  v_{2}=\frac{m_{1}v_{1}}{m_{2}}\\ v_{2}=\frac{(0.15kg)(32.0m/s)}{65.0kg}\\ v_{2}=0.0738m/s\\or\\v_{2}=0.07 m/s

So Option d is correct one

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For a projectile launched horizontally, which of the following best describes the downward component of a projectile's velocity?
Angelina_Jolie [31]

C. The downward component of the projectile's velocity continually increases

Explanation:

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- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction  

Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:

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This means that

C. The downward component of the projectile's velocity continually increases

Because every second, it increases by 9.8 m/s in the downward direction.

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8 0
3 years ago
A car accelerates from rest at a constant rate of 2m/s2 for 5s. what is the speed of a car at the end of that time?
Fiesta28 [93]
Good afternoon.


We have:

\mathsf{V_0 = 0}\\ \mathsf{a = 2 \ m/s^2}\\ \mathsf{t = 5 \ s}

The function of velocity:

\mathsf{V = V_0+at}\\ \\ \mathsf{V = 0 + 2t}\\ \\ \mathsf{V = 2t}


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3 years ago
Need a little help here :(
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Answer:

The output out be 200

Explanation:

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3 years ago
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On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the
IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

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(a) Energy density of electric field = \frac{1}{2}\varepsilon _{0}E^{2}

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(b) energy density of magnetic field = \frac{B^{2}}{2\mu _{0}}

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= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0
3 years ago
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