NEET-XII-Physics
42: Photoelectric Effect and Wave Particle Duality
- #3An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process.Ans : Given:
Wavelength of absorbed photon, `` {\lambda }_{1}`` = 500 nm
Wavelength of emitted photon, `` {\lambda }_{2}`` = 700 nm
Speed of light, c = 3`` \times {10}^{8}`` m/s
Planck's constant, h = 6.63`` \times ``10`` -34`` Js
Energy of absorbed photon,
`` {E}_{1}=\frac{hc}{{\lambda }_{1}}=\frac{h\times 3\times {10}^{8}}{500\times {10}^{-9}}``
Energy of emitted photon,
`` {E}_{2}=\frac{hc}{{\lambda }_{2}}=\frac{h\times 3\times {10}^{8}}{700\times {10}^{-9}}``
Energy absorbed by the atom in the process:
`` {E}_{1}-{E}_{2}=hc\left[\frac{1}{{\lambda }_{1}}-\frac{1}{{\lambda }_{2}}\right]``
`` =6.63\times 3\left[\frac{1}{5}-\frac{1}{7}\right]\times {10}^{-19}``
`` =6.63\times 3\times \frac{2}{35}\times {10}^{-19}``
`` =1.136\times {10}^{-19}\,\mathrm{\,J\,}``
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