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## QuestionPhysicsClass 12

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?
$$h=6.62\times 10^{-34}\;\text{Js}$$
$$e=1.6\times 10^{-19}\;\text{C}$$

$$\lambda _1=390\;\mathrm{nm}$$
$$\lambda _2=550\;\mathrm{nm}$$
$$\lambda _3=760\;\mathrm{nm}$$
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
$$E=\frac{\text{hc}}{\lambda }$$
$$E_1=\frac{\mathit{hc}}{\lambda _1}$$
$$=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{390\times 10^{-9}}$$
$$=5.10\times 10^{-19}\;\text{V}$$
$$=\frac{\left(5.10\times 10^{19}\right)}{1.6\times 10^{-19}}=3.19\;\text{eV}$$
$$E_2=\frac{\text{hc}}{\lambda _2}$$
$$=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{550\times 10^{-9}}$$
$$=3.62\times 10^{-19}\;\text{V}$$
$$=\frac{\left(3.62\times 10^{-19}\right)}{1.6\times 10^{-19}}=3.19\;\text{eV}$$
$$E_3=\frac{\mathit{hc}}{\lambda _3}$$
$$=\frac{\left(6.62\times 10^{-34}\times 3\times 10^8\right)}{760\times 10^{-9}}$$
$$=2.62\times 10^{-19}\;\text{V}$$
$$=\frac{\left(2.62\times 10^{-19}\right)}{1.6\times 10^{-19}}=1.64\;\text{eV}$$          