Solved Problem on Photons
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Light of wavelength 2000 Å falls on an aluminum surface. In aluminum 4.2 eV are required to remove an electron. What is the kinetic energy of
a) the fastest and
b) the slowest emitted photoelectrons;
c) What is the stopping potential?
d) What is the cutoff wavelength for aluminum?
e) If the intensity of the incident light is 2.0 W/m2, what is the average number of photons per unit time per unit area that strike the surface?
f) If the incident light intensity is doubled, what is the maximum energy value of the removed electrons?
Given: speed of light, c = 2.998×108 m/s, Planck constant, h = 6.626×10&mminus;34 J.s, and 1 eV = 1.602×10−19 J.
Problem data:
- Aluminum work function: ϕ = 4.2 eV;
- Wavelength of incident radiation: λ = 2000 Å;
- Incident light intensity: I= 2.0 W/m2;
- Speed of light: c = 2.998×108 m/s;
- Planck constant: h = 6.626×10−34 J.s;
- Electron volts: 1 eV = 1.602×10−19 J.
Solution
First, let's convert the given wavelength in angstroms (Å) to meters (m) and the work function given in electron volts (eV) to joules (J) used in the International System of Units (SI)
\[
\begin{gather}
\lambda=2000\;\cancel{\mathrm{\mathring{A}}}\times \frac{1\times 10^{-10}\;\text{m}}{1\;\cancel{\mathrm{\mathring{A}}}}=2\times 10^{3}\times 10^{-10}\;\text{m}=2.0\times 10^{-7}\;\text{m}\\[10pt]
\phi=4.2\;\cancel{\text{eV}}\times \frac{1.602\times 10^{-19}\;\text{J}}{1\;\cancel{\text{eV}}}=6.7\times 10^{-19}\;\text{J}
\end{gather}
\]
a) The kinetic energy (K) with which a photoelectron is ejected is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{K=h \nu-\phi} \tag{I}
\end{gather}
\]
The relationship between frequency and wavelength is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{c=\lambda\nu}
\end{gather}
\]
\[
\begin{gather}
\nu=\frac{c}{\lambda} \tag{II}
\end{gather}
\]
substituting relation (II) into equation (I)
\[
\begin{gather}
K_{max}=h\frac{c}{\lambda}-\phi \tag{III}
\end{gather}
\]
substituting the problem data, the kinetic energy of the fastest electron will be
\[
\begin{gather}
K_{max}=6.626\times 10^{-34}\times \frac{2.998\times 10^{8}}{2.0\times 10^{-7}}-6.7\times 10^{-19}\\[5pt]
K_{max}=9.9\times 10^{-19}-6.7\times 10^{-19}\\[5pt]
K_{max}=3.2\times 10^{-19}\;\text{J}
\end{gather}
\]
converting this energy to electron volts
\[
\begin{gather}
K_{max}=3.2\times 10^{-19}\;\cancel{\text{J}}\times \frac{1\;\text{eV}}{1.602\times 10^{-19}\;\cancel{\text{J}}}=2.0\;\text{eV}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{K_{max}=2.0\;\text{eV}}
\end{gather}
\]
b) The electrons are ejected with an energy distribution from zero to maximum energy
(Kmax), and the kinetic energy of the slowest electron will be
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{K_{max}=0}
\end{gather}
\]
c) The maximum kinetic energy as a function of the electron charge (e) and the potential difference (V0) to which the electrons are subjected
\[
\begin{gather}
\bbox[#99CCFF,10px]
{K_{max}=eV_{0}}
\end{gather}
\]
\[
\begin{gather}
V_{0}=\frac{K_{max}}{e}\\[5pt]
V_{0}=\frac{2,0}{1}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{V_{0}=2.0\;\text{V}}
\end{gather}
\]
d) The cutoff wavelength occurs when the kinetic energy is equal to zero (K = 0), using equation (III)
\[
\begin{gather}
K=h\frac{c}{\lambda}-\phi\\[5pt]
0=6.62\times 10^{-34}\times \frac{2.998\times 10^{8}}{\lambda}-6.7\times 10^{-19}\\[5pt]
\frac{1.4\times 10^{-24}}{\lambda}=6.7\times 10^{-19}\\[5pt]
\lambda=\frac{1.4\times 10^{-24}}{6.7\times 10^{-19}}\\[5pt]
\lambda=2.957\times 10^{-7}\;\text{m}
\end{gather}
\]
converting this wavelength to angstroms
\[
\begin{gather}
\lambda=2.957\times 10^{-7}\;\cancel{\text{m}}\times \frac{1\;\mathrm{\mathring{A}}}{1\times 10^{-10}\;\cancel{\text{m}}}=2.957\times 10^{-7}\times 1\times 10^{10}\;\mathrm{\mathring{A}}=2957\;\mathrm{\mathring{A}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\lambda=2957\;\mathrm{\mathring{A}}}
\end{gather}
\]
e) The energy of a photon is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{E=h \nu} \tag{IV}
\end{gather}
\]
substituting relation (II) into equation (IV)
\[
\begin{gather}
E=h\frac{c}{\lambda}\\[5pt]
E=6.62\times 10^{-34}\times \frac{2.998\times 10^{8}}{2\times 10^{-7}}\\[5pt]
E\approx 1.0\times 10^{-18}\;\frac{\text{J}}{\text{photon}}
\end{gather}
\]
The intensity (I) of the incident light is given by the power
(\(\mathscr{P}\))
per unit area (A)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{I=\frac{\mathscr{P}}{A}} \tag{V}
\end{gather}
\]
Power (\( \mathscr{P} \)) is given by energy (E) per unit time (Δt)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathscr{P}=\frac{E}{\Delta t}} \tag{VI}
\end{gather}
\]
substituting equation (VI) into equation (V)
\[
\begin{gather}
I=\frac{E}{\Delta t A}
\end{gather}
\]
The number of photons (n) is obtained by dividing the light intensity by the photon energy
\[
\begin{gather}
n=\frac{I}{E}\\[5pt]
n=\frac{2.0\;\frac{\cancel{\text{J}}}{\text{s.m}^{2}}}{1.0\times 10^{-18}\;\frac{\;\cancel{\text{J}}}{\text{photons}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{n=2.0\times 10^{18}\;\frac{\text{photons}}{\text{s.m}^{2}}}
\end{gather}
\]
f) The maximum energy of the electrons will remain the same, and the energy depends on the frequency of the incident light (equation IV), not on the intensity.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .