Physics, 19.07.2019 16:00, 19colemankl

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

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answered: escojo942

The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

Further explanation

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

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

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

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.

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

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

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem!

Given:

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

Unknown:

fo = ?

Solution:

$\Phi = h \times fo$

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

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

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

Learn morePhotoelectric Effect : Statements about the Photoelectric Effect : Rutherford model and Photoelecric Effect : Photoelectric Threshold Wavelength : Answer details

Grade: High School

Subject: Physics

Chapter: Quantum Physics

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

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

answered: chanel2371

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, 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 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 in expression for work function.

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