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Physics Planck Equation Question
(c) Tenaga minimum yang diperlukan untuk fotoelektron terlepas dari permukaan logam litium ialah 2.32 \mathrm{eV}.
The minimum energy required for a photoelectron to escape from the lithium metal surface is 2.32 \mathrm{eV}.(i) Logam litium tidak menunjukkan kesan fotoelektrik untuk cahaya biru yang mempunyai panjang gelombang 460 \mathrm{~nm}. Terangkan. Lithium metal does not show a photoelectric effect for blue light with a wavelength of 460 \mathrm{~nm}. Explain.
(ii) Berapakah panjang gelombang ambang bagi litium? What is the threshold wavelength for lithium?
Physics Planck Equation Solution
Part c(ii)
First, we need to identify this question is from which Physics topic. Looking at keywords like “photoelectric”, “energy” and “wavelength”, we know it is certainly related to Planck’s equation.
Now let’s refer to Planck’s equation definition:
The Planck relation states that the energy of a photon, E, known as photon energy, is proportional to its frequency, ν: The constant of proportionality, h, is known as the Planck constant.
Its formula is:
E = hv = \frac{hc}{\lambda}
where h = Planck Constant, c = speed of light, \lambda = wavelength, E=energy
From the question, we can extract following information:
Minimum Energy for a photoelectron to escape, E = 2.32 eV
Note 1eV = 1.6 \times 10^{-19}V,
Hence, E = 2.32 eV = 2.32 \times 1.6 \times 10^{-19}
From formula sheet,
Planck Constant, h = 6.626 \times 10^{-34}
Speed of light, c = 3.0 \times 10^8
Plugging all the above values into Planck’s equation,
(2.32 \times 1.6 \times 10^{-19}) = \frac{(6.626 \times 10^{-34}) \times ( 3.0 \times 10^8)}{\lambda}
\lambda = 5.355 \times 10^{-7} m
\lambda = 535.5 \times 10^{-9} m
\lambda = 535.5 nm
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