A) 2 eV
B) 1 eV
C) 4 eV
D) 0.2 eV
Correct Answer: A
Solution :
[a] By Einstein's equation of photo-electric effect, the maximum kinetic energy of emitted photo-electrons is given by \[{{E}_{k}}=hv-W\]or \[{{E}_{k}}=\frac{hc}{\lambda }-W\] Where, h = Planck's constant v = frequency of incident light W= work function ofmetal \[\lambda \]= wavelength of incident light \[{{E}_{k}}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2000\times {{10}^{-10}}\times 1.6\times {{10}^{-19}}}eV-4.2eV\] So, \[{{E}_{k}}=2eV\]
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