A) 494 nm
B) 594
C) 640 nm
D) 700 nm
Correct Answer: A
Solution :
\[E=\frac{hc}{\lambda }\] \[=\frac{242\times {{10}^{+3}}}{6.02\times {{10}^{23}}}=\frac{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}{\lambda }\] \[\lambda =\frac{6.62\times {{10}^{-26}}\times 3\times 6.02\times {{10}^{20}}}{242}\] \[=\frac{6.62\times 18.06\times {{10}^{-6}}}{242}\] \[=0.494\times {{10}^{-6}}=4.94\times 10{{\text{ }}^{-7}}m\] \[=494\text{ }nm\]You need to login to perform this action.
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