Question

What is the threshold frequency ν0 of cesium?

Answer 1

You are given a ν0 of cesium and you are asked to find the threshold frequency. First, you have to know the half reaction of cesium which is

Cs(g) → Cs⁺ + e⁻ with a ΔHIP of 375.7 kJ/mol
Then convert kilojoules to joules and then divide this by the Avogadro's constant which is 6.023 x 10²³ atoms/mole to give the energy of photon in joules per photon

E = (375,700 J/mol) / (6.023 x 10²³ photons/ mole)
E = 6.239 x 10⁻¹⁹ J/photon

Use the equation of energy of a photon
E = hv where E is the energy of the photon, h is the planck's constant and v is the threshold frequency
E = hv
6.239 x 10⁻¹⁹ J/photon = 6.626 x 10⁻³⁴ (v)
v = 9.42 x 10¹⁴ s⁻¹ Hz

Answer 2

The threshold frequency fo of Cesium is about 9.39 × 10¹⁴ Hz

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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 )

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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!

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Complete Question:

What is the threshold frequency fo of Cesium?

Note that 1 eV (electron volt) = 1.60×10⁻¹⁹ J ?

$\begin{tabular} {|c|c|c|}\underline{Light\ Energy\ (eV)}& \underline{Electron\ Emitted} & \underline{Electron KE\ (eV)}\\3.87 & no & -\\3.88 & no & -\\ \boxed{3.89} & yes & 0\\3.90 & yes & 0.01\\3.91 & yes & 0.02\end{tabular}$

Given:

the work function = Φ = 3.89 eV

Asked:

threshold frequency = fo = ?

Solution:

Firstly, we will convert the work function unit as follows:

Φ = 3.89 eV = 3.89 × 1.60 × 10⁻¹⁹ Joule = 6.224 × 10⁻¹⁹ Joule

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Next, we will calculate the threshold frequensy as follows:

$\Phi = h \times fo$

$6.224 \times 10^{-19} = 6.63 \times 10^{-34} \times fo$

$fo = ( 6.224 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )$

${\boxed{fo \approx 9.39 \times 10^{14} ~ Hz}$

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Learn more

• 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

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Answer details

Grade: High School

Subject: Physics

Chapter: Quantum Physics