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

Answer:

$\large \boxed{\text{1.016 \times 10^{15} Hz}}$

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

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