A) 6.56 nm
B) 65.6 nm
C) 656 nm
D) 0,656 nm (Given, h(Planck's constant) \[=6.63\,\times {{10}^{-34}}\,J-s,\] c(velocity of light) \[=3.00\,\times {{10}^{8}}\,m{{s}^{-1}}\])
Correct Answer: C
Solution :
According to formula, \[E=\frac{hc}{\lambda }\] \[3.03\times {{10}^{-19}}=\frac{hc}{\lambda }\] \[\lambda =\frac{6.63\times {{10}^{-34}}\times 3.00\times {{10}^{8}}}{3.03\times {{10}^{-19}}}\] \[=6.56\times {{10}^{-7}}\,m\] \[=6.56\times {{10}^{-7}}\times {{10}^{9}}\,\text{nm}\] \[=6.56\times {{10}^{2}}\,nm=656\,nm\]You need to login to perform this action.
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