A) \[1.24\times {{10}^{12}}{{s}^{-1}}\]
B) \[3.15\times {{10}^{14}}{{s}^{-1}}\]
C) \[4.76\times {{10}^{14}}{{s}^{-1}}\]
D) \[1.24\times {{10}^{15}}{{s}^{-1}}\]
Correct Answer: D
Solution :
\[E=hv\], hence, \[v=\frac{E}{h}\]\[=\frac{495.8\times {{10}^{3}}J/\text{atom}}{6.02\times {{10}^{23}}\times 6.62\times {{10}^{-34}}Js}\]\[=1.24\times {{10}^{15}}{{s}^{-1}}\]
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