The values of Planck's constant is $$6.63 \times {10^{ - 34}}\,Js.$$   The velocity of light is $$3.0 \times {10^8}m\,{s^{ - 1}}.$$   Which value is closest to the wavelength in nanometres of a quantum of light with frequency of $$8 \times {10^{15}}{s^{ - 1}}?$$

Solution :
$$\eqalign{ & E = h\nu = \frac{{ch}}{\lambda };\,{\text{and}}\,\nu = \frac{c}{\lambda } \cr & 8 \times {10^{15}} = \frac{{3.0 \times {{10}^8}}}{\lambda } \cr & \therefore \,\,\lambda = \frac{{3.0 \times {{10}^8}}}{{8 \times {{10}^{15}}}} \cr & = 0.37 \times {10^{ - 7}} \cr & = 37.5 \times {10^{ - 9}}m \cr & = 4 \times {10^1}nm \cr} $$