Solved Problem on Quantum Physics
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A cesium photoelectron has a kinetic energy of 2 eV.
a) What is the maximum frequency of light that could have emitted this electron? Data: Cesium work function = 1.8 eV, Planck's constant h = 6.62×10−34 J.s, 1 eV = 1.6×10−19 J.
b) What is the maximum wavelength of this frequency? Data: speed of light c = 3×108 m/s.


Problem data:
  • Cesium photoelectron kinetic energy:    Ec = 2 eV;
  • Cesium work function:    φ = 1.8 eV;
  • Planck's constant:    h = 6.62 ×10−34 J.s;
  • Conversion factor of electron-volt/Joule:    1 eV = 1.6 ×10−19 J;
  • Speed ​​of light:    c = 3 ×108 m/s.
Problem diagram:

Radiation of frequency f and energy hf incident on a cesium surface that emits an electron with energy K (Figure 1).
Figure 1

Solution

First, we must convert the kinetic energy and work function values ​​given in electron volts (eV) to joules (J), to make it compatible with the value of Planck's Constant h given in joules.second (J.s).
For kinetic energy using a cross-multiplication
\[ \begin{gather} \frac{1\;\text{eV}}{1.6\times 10^{-19}\;\text{J}}=\frac{2\;\text{eV}}{K}\\[5pt] K=\frac{2\;\cancel{\text{eV}}\times 1.6\times 10^{-19}\;\text{J}}{1\;\cancel{\text{eV}}}\\[5pt] K=3.2\times 10^{-19}\;\text{J} \end{gather} \]
For the work function
\[ \begin{gather} \frac{1\;\text{eV}}{1.6\times 10^{-19}\;\text{J}}=\frac{1.8\;\text{eV}}{\phi}\\[5pt] \phi=\frac{1.8\;\cancel{\text{eV}}\times 1.6\times 10^{-19}\;\text{J}}{1\;\cancel{\text{eV}}}\\[5pt] \phi=2.88\times 10^{-19}\;\text{J} \end{gather} \]
a) For the given kinetic energy the maximum frequency will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {hf=\phi +E_{c}} \end{gather} \]
\[ \begin{gather} 6.62\times 10^{-34}f=(2.88-3.2)\times 10^{-19}\\[5pt] f=\frac{6,08.10^{-19}}{6.62\times 10^{-34}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {f=9.2\times 10^{14}\;\text{Hz}} \end{gather} \]

b) Frequency and wavelength are related by
\[ \begin{gather} \bbox[#99CCFF,10px] {c=\lambda f} \end{gather} \]
\[ \begin{gather} \lambda =\frac{c}{f} \end{gather} \]
substituting the speed of light c given in the statement and the frequency f found in the previous item
\[ \begin{gather} \lambda =\frac{3\times 10^{8}}{9.2\times 10^{14}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\lambda =3.3\times10^{7}\;\text{m}} \end{gather} \]
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