A) \[1.7\,\,eV\]
B) \[1.6\,\,eV\]
C) \[1.5\,\,eV\]
D) \[1.0\,\,eV\]
Correct Answer: D
Solution :
If the maximum kinetic energy of photoelectron emitted from the surface of a metal is\[{{E}_{K}}\]and\[W\]is the work function of the metal, the from Einsteins photoelectric equation we have \[{{E}_{K}}=hv-W\] where,\[hv\]is the energy of the photon absorbed by the electron in the metal. Also \[v=\frac{C}{\lambda }\] \[=\frac{velocity}{wavelength}\] \[={{E}_{K}}=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}=hc\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] Given, \[\lambda =3\times {{10}^{-7}}m\] \[{{\lambda }_{0}}=4\times {{10}^{-7}}m\] \[\therefore \] \[{{E}_{K}}=6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}\] \[\times \left( \frac{1}{3\times {{10}^{-7}}}-\frac{1}{4\times {{10}^{-7}}} \right)J\] \[=\frac{19.8\times {{10}^{-19}}}{12\times 1.6\times {{10}^{-19}}}eV\] \[=1.03\,\approx 1eV\]
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