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प्रश्न
Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
उत्तर
Given :
Range of wavelengths, `λ_1` = 400 nm to `λ_2` = 780 nm
Planck's constant, `h = 6.63 xx 10^-34 "Js"`
Speed of light, `c = 3 xx 10^8 "m/s"`
Energy of photon,
`E = hv`
`v = c/λ`
`therefore E = hv = (hc)/λ`
Energy `(E_1)` of a photon of wavelength `(λ_1)` :
`E_1 = (hc)/λ_1`
= `(6.63 xx 10^-34 xx 3 xx 10^8)/(400 xx 10^-9)`
= `(6.63 xx 3)/4 xx 10^-9`
= `4.97725 xx 10^-19`
= `5 xx 10^-19 "J"`
Energy `(E_2)` of a photon of wavelength `(λ_2)` :
`E_2 = (6.63 xx 3)/7.8 xx 10^-19`
= `2.55 xx 10^-9 "J"`
So, the range of energy is `2.55 xx 10^-19 "J"` to `5 xx 10^-19 "J"` .
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