The wavelength of light in the visible region is about \(390\;\mathrm{mm}\) for violet colour, about \(550\;\mathrm{mm}\) \((\)average wavelength\()\) for yellow-green color and about \(770\;\mathrm{nm}\) for red colour.
What are the energies of photons in \(\;\text{eV}\) at the \((i)\) violet end, \((\mathit{ii})\) average wave length, yellow-green colour, and \((iii)\) red end of the visible spectrum?
\(h=6.62\times 10^{-34}\;\text{Js}\)
\(e=1.6\times 10^{-19}\;\text{C}\)
What are the energies of photons in \(\;\text{eV}\) at the \((i)\) violet end, \((\mathit{ii})\) average wave length, yellow-green colour, and \((iii)\) red end of the visible spectrum?
\(h=6.62\times 10^{-34}\;\text{Js}\)
\(e=1.6\times 10^{-19}\;\text{C}\)
Answer
\(\lambda _1=390\;\mathrm{nm}\)
\(\lambda _2=550\;\mathrm{nm}\)
\(\lambda _3=760\;\mathrm{nm}\)
Respectively for violet, yellow-green and red colour.
Consider \(E_1,E_2\;\text{ and}\;E_3\) to be their energies respectively.
Now by using the following relation
\(E=\frac{\text{hc}}{\lambda }\)
\(E_1=\frac{\mathit{hc}}{\lambda _1}\)
\(=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{390\times 10^{-9}}\)
\(=5.10\times 10^{-19}\;\text{V}\)
\(=\frac{\left(5.10\times 10^{19}\right)}{1.6\times 10^{-19}}=3.19\;\text{eV}\)
\(E_2=\frac{\text{hc}}{\lambda _2}\)
\(=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{550\times 10^{-9}}\)
\(=3.62\times 10^{-19}\;\text{V}\)
\(=\frac{\left(3.62\times 10^{-19}\right)}{1.6\times 10^{-19}}=3.19\;\text{eV}\)
\(E_3=\frac{\mathit{hc}}{\lambda _3}\)
\(=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{760\times 10^{-9}}\)
\(=2.62\times 10^{-19}\;\text{V}\)
\(=\frac{\left(2.62\times 10^{-19}\right)}{1.6\times 10^{-19}}=1.64\;\text{eV}\)
\(\lambda _2=550\;\mathrm{nm}\)
\(\lambda _3=760\;\mathrm{nm}\)
Respectively for violet, yellow-green and red colour.
Consider \(E_1,E_2\;\text{ and}\;E_3\) to be their energies respectively.
Now by using the following relation
\(E=\frac{\text{hc}}{\lambda }\)
\(E_1=\frac{\mathit{hc}}{\lambda _1}\)
\(=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{390\times 10^{-9}}\)
\(=5.10\times 10^{-19}\;\text{V}\)
\(=\frac{\left(5.10\times 10^{19}\right)}{1.6\times 10^{-19}}=3.19\;\text{eV}\)
\(E_2=\frac{\text{hc}}{\lambda _2}\)
\(=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{550\times 10^{-9}}\)
\(=3.62\times 10^{-19}\;\text{V}\)
\(=\frac{\left(3.62\times 10^{-19}\right)}{1.6\times 10^{-19}}=3.19\;\text{eV}\)
\(E_3=\frac{\mathit{hc}}{\lambda _3}\)
\(=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{760\times 10^{-9}}\)
\(=2.62\times 10^{-19}\;\text{V}\)
\(=\frac{\left(2.62\times 10^{-19}\right)}{1.6\times 10^{-19}}=1.64\;\text{eV}\)
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