Question

The wavelength of red light is about $\mathbf{695}\mathbf{}\mathit{n}\mathit{m}$. What are the frequency and period of this radiation?

Short answer

The frequency of the red light and period of radiation are $4.32\times {10}^{14}\text{Hz}$ and$23.1\times {10}^{-16}\text{s}$ respectively

Step-by-step solution

Identification of given data

The wavelength of red light, $\lambda =695\text{nm}$.

Determine the formulas to calculate the frequency of the red light and period of radiation.

The expression to calculate the frequency is given as follows.

$\mathit{f}\mathbf{=}\frac{\mathbf{v}}{\mathbf{\lambda }}$ …(i)

Here, $v$is the speed of the light.

The expression to calculate the period of the radiation is given as follows.

role="math" localid="1668595348890" $\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$ … (ii)

Determining the frequency of the red light and period of radiation.

$f$Calculate the frequency of the red light.

Substitute$3\times {10}^{8}m/s$ for$\nu$ and$695\text{nm}$ for$\lambda$ into equation (i).

$$\begin{gathered} f=\frac{3\times {10}^8 m/s}{695\times {10}^{-9}} \\ =\frac{3000\times {10}^{5}m/s }{695\times {10}^{-9} \text{m}} \\ =\frac{3000\times {10}^{14}m/s }{695 \text{m}} \\ =4.32\times {10}^{14}\text{Hz} \end{gathered}$$

Calculate the period of radiation.

Substitute$4.32\times {10}^{14}\text{Hz}$ for into equation (ii).

$$\begin{gathered} T=\frac{1}{4.32\times {10}^{14} \text{Hz}} \\ =0.231\times {10}^{-14}\text{s} \\ =23.1\times {10}^{-16}\text{s} \end{gathered}$$

Hence the frequency of the red light and period of radiation are$4.32\times {10}^{14}\text{Hz}$ and $23.1\times {10}^{-16}\text{s}$respectively.