• # question_answer Calculate the energy in joule corresponding to light of wavelength 45 nm: (Planck's constant $h=6.63\times {{10}^{-34}}\,Js;$speed of light$c=3\times {{10}^{8}}\,m{{s}^{-1}}$) A) $6.67\times {{10}^{15}}$B) $6.67\times {{10}^{11}}$C) $4.42\times {{10}^{-15}}$D) $4.42\times {{10}^{-18}}$

$\frac{hc}{\lambda }=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{45\times {{10}^{-9}}}=4.42\times {{10}^{-18}}J$