Question

This shielding is a maximum for UV light having a wavelength of 295nanometers. What is the frequency in hertz of this particular wavelength of UV light

Answer 1

`\large \boxed{\text{1.016 $* 10^(15)$ Hz}}`

Step-by-step explanation:

The formula relating the frequency (f) and wavelength (λ) is

fλ = c

1. Convert nanometres to metres

295 nm = 295 × 10⁻⁹ m

2. Calculate the frequency

`\begin{array}{rcl}f * 295 * 10^(-9) \text{ m}& =& 2.998 * 10^(8) \text{ m$\cdot$s$^(-1)$}\\\\f & = & \frac{2.998 * 10^(8) \text{ m$\cdot$s$^(-1)$}}{295 * 10^(-9)\text{ m}}\\\\& = &1.016 * 10^(15)\text{ s}^(-1)\\& = &\large \boxed{\textbf{1.016 $\mathbf{* 10^(15)}$ Hz}}\end{array}`