Discussion Forum

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?

Answer:

Explanation:

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