A) \[39\,\,eV\]
B) \[27\,\,eV\]
C) \[54\,\,eV\]
D) \[13.5\,\,eV\]
Correct Answer: B
Solution :
From Einstein?s photoelectric equation, if the kinetic energy of photoelectrons emitted from metal surface is\[{{E}_{K}}\], and W is the work function of metal. Then \[{{E}_{K}}=hv-W\] Where \[hv\] is energy of photon absorbed by the electron in the metal. Given, \[v=8\times {{10}^{15}}\,\,Hz,\,h=6.6\times {{10}^{-34}}\,J-s\] Also \[1eV=1.6\times {{10}^{-19}}\,J\] \[W=6.125\times 1.6\times {{10}^{-19}}V\] \[=9.8\times {{10}^{-19}}V\] \[\therefore {{E}_{K}}=6.6\times {{10}^{-34}}\times 8\times {{10}^{15}}-9.8\times {{10}^{-19}}\] \[=52.8\times {{10}^{-19}}-9.8\times {{10}^{-19}}\] \[=43\times {{10}^{-19}}J\] Also, \[1\,eV=1.6\times {{10}^{-19}}\,J\] \[\therefore \] \[{{E}_{K}}=\frac{43\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=26.87\,\,eV\] \[{{E}_{K}}\approx \,27\,\,eV\] Note: when photon falls on a metal, it transfers whole of its energy to one of the electrons present and its own existence is vanished. The electrons emitted from the surface of a metal have maximum kinetic energy, because their energy is not lost by collisions.
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