Solved Problem on Photons

a) The energy required to remove an electron from sodium is 2.3 eV. Does sodium show a photoelectric effect for the yellow light, with λ = 5890 Å?
b) What is the cutoff wavelength for photoelectric emission from sodium?
Given: speed of light, c = 2.998×10⁸ m/s, Planck constant, h = 6.626×10⁻³⁴ J.s, and 1 eV = 1.602×10⁻¹⁹ J.

Problem data:

Sodium work function: ϕ = 2.3 eV;
Incident light wavelength: λ = 5890 Å;
Speed of light: c = 2.998×10⁸ m/s
Planck constant:: h = 6.626×10⁻³⁴ J.s;
Electron-volt: 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)

\[ \begin{gather} \lambda=5890\;\cancel{\mathrm{\mathring{A}}}\times\frac{1\times 10^{-10}\;\text{m}}{1\;\cancel{\mathrm{\mathring{A}}}}=5.890\times 10^{3}\times 10^{-10}\;\text{m}=5.890\times 10^{-7}\;\text{m}\\[10pt] \phi=2.3\;\cancel{\text{eV}}\times\frac{1.602\times10^{-19}\;\text{J}}{1\;\cancel{\text{eV}}}=3.695\times 10^{-19}\;\text{J} \end{gather} \]

a) The energy of a photon as a function of frequency is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {E=h\nu} \tag{I} \end{gather} \]

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)

\[ \begin{gather} E=h\frac{c}{\lambda}\\[5pt] E=6.626\times 10^{-34}\times \frac{2.998\times 10^{8}}{5.89\times 10^{-7}}\\[5pt] E=3.372\times 10^{-19}\;\text{J} \end{gather} \]

converting this energy to electron volts

\[ \begin{gather} E=3.372\times 10^{-19}\;\cancel{\text{J}}\times \frac{1\;\text{eV}}{1.602\times 10^{-19}\;\cancel{\text{J}}}=2.1\;\text{eV} \end{gather} \]

The photon energy is less than the work function to remove an electron from sodium (E < ϕ), and sodium has no photoelectric effect for the wavelength of 5890 Å.

b) The kinetic energy (K) with which emits a photoelectron is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {K=h\nu-\phi} \tag{III} \end{gather} \]

The cutoff wavelength occurs when the kinetic energy is equal to zero (K = 0), substituting the work function given in the problem and the relation (II) into equation (III)

\[ \begin{gather} 0=h\frac{c}{\lambda}-\phi\\[5pt] 3.695\times 10^{-19}=6.626\times 10^{-34}\times \frac{2.998\times 10^{8}}{\lambda}\\[5pt] \lambda=6.626\times 10^{-34}\times \frac{2.998\times 10^{8}}{3.695\times 10^{-19}}\\[5pt] \lambda=5.376\times 10^{-7}\approx5.4\times 10^{-7}\;\text{m} \end{gather} \]

converting this wavelength to angstroms

\[ \begin{gather} \lambda=5.4\times 10^{-7}\;\cancel{\text{m}}\times \frac{1\;\mathrm{\mathring{A}}}{1\times 10^{-10}\;\cancel{\text{m}}}=5.4\times 10^{-7}\times 10^{10}\;\mathrm{\mathring{A}}=5400\;\mathrm{\mathring{A}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\lambda=5400\;\mathrm{\mathring{A}}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .