A) 1.6
B) 2.5
C) 4.1
D) 6.6
Correct Answer: A
Solution :
Given, frequency of light \[={{10}^{15}}Hz\] Work function = 2.5 eV Now, wavelength of incident light \[\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{{{10}^{15}}}=3\times {{10}^{-7}}\] \[=3000\,\,\overset{o}{\mathop{A}}\,\] The energy of \[=3000\,\,\overset{o}{\mathop{A}}\,\] light \[=\frac{hc}{\lambda }\] \[=\frac{12400}{3000}=4.13\,eV\] \[\therefore \] Maximum kinetic energy of the electron \[=hv-W\] and \[{{K}_{\max }}=e{{V}_{0}}\] \[\therefore \] \[1.63\,eV\,=e{{V}_{0}}\] \[\Rightarrow \] \[{{V}_{0}}=1.63\,V\]
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