The values of Planck's constant is $6.63 \times 10^{-34} \mathrm{Js}$. The velocity of light is $3.0 \times…
the wavelength in nanometres of a quantum of light with frequency of $8 \times 10^{15} \mathrm{~s}^{-1}$ ?
- $5 \times 10^{-18}$
- $4 \times 10^{1}$
- $3 \times 10^{7}$
- $2 \times 10^{-25}$
Solution
$8 \times 10^{15}=\frac{3.0 \times 10^{8}}{\lambda}$
$\therefore \lambda=\frac{3.0 \times 10^{8}}{8 \times 10^{15}}=0.37 \times 10^{-7}=37.5 \times 10^{-9} \mathrm{~m}=4 \times 10^{1}$ .
Asked in: JEE-TOPICTESTS-CHEMISTRY