1) The values of Planck's constant is 6.63 x 1034 Js. The velocity of light is 3.0 x 108 m s-1. Which value is closest to the wavelength in nanometres of a quantum of light with a frequency of 8 x 1015 s-1? A) $5\times10^{-18}$ B) $4\times10^{1}$ C) $3\times10^{7}$ D) $2\times10^{-25}$ Answer: Option BExplanation:E = hν = $\frac{Ch}{\lambda};and\upsilon = \frac{c}{\lambda}$ 8×1015 = $\frac{3.0\times10^{8}}{\lambda}$ $\therefore\lambda=\frac{3.0\times10^{8}}{8\times10^{15}}=0.37\times10^{-7}$ =37.5x10-9 m = 4 x 101 nm
