Question
Download Solution PDFThe work function of a photometal is 6.63 eV. The threshold wavelength is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
Work function (Φo):
- The minimum energy of incident radiation, required to eject the electrons from the metallic surface is defined as the work function of that surface.
\(\Rightarrow {{\rm{\Phi }}_0} = h{\nu _0} = \frac{{hc}}{{{\lambda _0}}}\)
Where h = Plank's constant, c = speed of light, ν0 = threshold frequency and λ0 = threshold wavelength.
EXPLANATION:
Given - Work function (Φo) = 6.63 eV
- The threshold wavelength is
\(\Rightarrow \Phi_o=\frac{12375}{\lambda_o(A^o)}eV\)
\(\Rightarrow \lambda_o=\frac{12375}{\Phi_o(A^o)}=\frac{12375}{6.63}\, A^o=1866\, A^o\)
Last updated on May 6, 2025
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