A) 540 nm
B) 400 nm
C) 310 nm
D) 220 nm
Correct Answer: C
Solution :
Given, \[W=4eV\] \[=4\times 1.6\times {{10}^{-19}}J\] \[=6.4\times {{10}^{-19}}J\] \[W=\frac{hc}{{{\lambda }_{\max }}}\] \[\Rightarrow \] \[{{\lambda }_{\max }}=\frac{hc}{W}\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6.4\times {{10}^{-19}}}\] \[=310nm\]
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