NEET-XII-Physics
42: Photoelectric Effect and Wave Particle Duality
- #1Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.Ans : Given:
Range of wavelengths, `` {\lambda }_{1}`` = 400 nm to `` {\lambda }_{2}`` = 780 nm
Planck's constant, h = 6.63`` \times ``10`` -``34 Js
Speed of light, c = 3`` \times ``108 m/s
Energy of photon,
`` E=hv``
`` \nu =\frac{c}{\lambda }``
`` \therefore E=h\nu =\frac{hc}{\lambda }``
Energy `` \left({E}_{1}\right)`` of a photon of wavelength `` \left({\lambda }_{1}\right)``:
`` ``
`` {E}_{1}=\frac{hc}{{\lambda }_{1}}``
`` =\frac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{400\times {10}^{-9}}``
`` =\frac{6.63\times 3}{4}\times {10}^{-9}``
`` =4.97725\times {10}^{-19}``
`` =5\times {10}^{-19}\,\mathrm{\,J\,}``
`` ``
Energy (E2) of a photon of wavelength (`` {\lambda }_{2}``):
`` ``
`` {E}_{2}=\frac{6.63\times 3}{7.8}\times {10}^{-19}``
`` =2.55\times {10}^{-9}\,\mathrm{\,J\,}``
So, the range of energy is 2.55 × 10-19 J to 5 × 10-19 J.
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