A) \[0.1\,eV\]
B) \[2.5\,eV\]
C) \[0.58\,eV\]
D) \[1.581\,eV\]
Correct Answer: C
Solution :
From Einstein's photo electric equation \[{{E}_{K}}=E-W\] Where \[{{E}_{K}}\]is the kinetic energy of emitted photoelectrons, W the work function and the energy supplied. \[E=h\upsilon =\frac{hc}{\lambda }\] Where h is Planck's constant, c is the speed of light and \[\lambda \] the wavelength. \[\therefore \] \[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}\] \[=3.96\times {{10}^{-19}}J\] Also, \[1eV=1.6\times {{10}^{-19}}J\] So, \[e=\frac{3.96\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=2.48eV\] Hence, \[{{E}_{K}}=2.48-1.90=0.58eV.\]
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