A) 1.4 eV
B) 4.9 eV
C) 3.1 eV
D) 1.6 eV
Correct Answer: C
Solution :
\[E=\frac{hc\,}{\lambda }\,\Rightarrow \,\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\,\Rightarrow \,\frac{3.32\times {{10}^{-19}}}{{{E}_{2}}}=\frac{4000}{6000}\] \[\Rightarrow \,\,{{E}_{2}}=4.98\times {{10}^{-19}}J=3.1\,eV.\]You need to login to perform this action.
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