Question

calculate the maximum wavelength, \\( \lambda_{\text{max}} \\), of electromagnetic radiation that could eject electrons from the surface of copper, which has a work function of \\( 7.26 \times 10^{-19} \\, \text{j} \\).
\\( \lambda_{\text{max}} = \\) m
if the maximum speed of the emitted photoelectrons is \\( 5.34 \times 10^{6} \\, \text{m/s} \\), what wavelength of electromagnetic radiation struck the surface and caused the ejection of the photoelectrons?
\\( \lambda = \\) m

Explanation:

Step1: Calculate \(\lambda_{max}\)

The work - function \(\phi\) is related to the energy of a photon \(E = h
u=\frac{hc}{\lambda}\). At the threshold (maximum wavelength \(\lambda_{max}\)), the energy of the photon is equal to the work - function \(\phi\).
We know that \(E=\frac{hc}{\lambda_{max}}\), where \(h = 6.626\times10^{-34}\space J\cdot s\) and \(c=3\times 10^{8}\space m/s\).
So, \(\lambda_{max}=\frac{hc}{\phi}\)
Substitute \(h = 6.626\times 10^{-34}\space J\cdot s\), \(c = 3\times 10^{8}\space m/s\) and \(\phi=7.26\times 10^{-19}\space J\)

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Step2: Calculate \(\lambda\) using the photoelectric effect equation

The photoelectric effect equation is \(E=\phi+\frac{1}{2}mv^{2}\), where \(m = 9.11\times 10^{-31}\space kg\) (mass of an electron)
First, find \(E\):
\(E=\phi+\frac{1}{2}mv^{2}\)
Substitute \(\phi = 7.26\times 10^{-19}\space J\), \(m = 9.11\times 10^{-31}\space kg\) and \(v = 5.34\times 10^{6}\space m/s\)

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\(E=7.26\times 10^{-19}+ 1.299\times 10^{-17}=1.3716\times 10^{-17}\space J\)
Since \(E=\frac{hc}{\lambda}\), then \(\lambda=\frac{hc}{E}\)
Substitute \(h = 6.626\times 10^{-34}\space J\cdot s\), \(c = 3\times 10^{8}\space m/s\) and \(E = 1.3716\times 10^{-17}\space J\)

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Answer:

\(\lambda_{max}=2.74\times 10^{-7}\space m\)
\(\lambda = 1.45\times 10^{-8}\space m\)