A) 6.56 nm
B) 65.6 nm
C) 656 nm
D) 0.656 nm
Correct Answer: C
Solution :
| According to formula, \[E=\frac{hc}{\lambda }\] |
| \[3.03\times {{10}^{-19}}=\frac{hc}{\lambda }\] |
| \[\lambda =\frac{6.63\times {{10}^{-34}}\times 3.00\times {{10}^{8}}}{3.03\times {{10}^{-19}}}\] |
| \[=6.56\times {{10}^{-7}}\,m\] |
| \[=6.56\times {{10}^{-7}}\times {{10}^{9}}\,\text{nm}\] |
| \[=6.56\times {{10}^{2}}\,nm=656\,nm\] |
| (Given, h(Planck's constant) \[=6.63\times {{10}^{-34}}~J-s,\](velocity of light)\[=3.00\times {{10}^{8}}m{{s}^{-1}}\]) |
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