Chapter 6, Problem 2SEPP

Interpretation Introduction

Interpretation:

The frequency of the sun is to be evaluated.

Concept introduction:

The relationship between wavelength and frequency of a radiation is as follows:

$\text{c}=\lambda \times \nu$

Here, $\text{c}$ is the speed of light $\left(3.0\times {10}^8\text{ m/s}\right)$, $\lambda$ is the wavelength, and $v$ is the frequency.

The most intense emission, $\left({\lambda }_{\max }\right)$ from sun occurs at $500\text{ nm}$.

To convert meter to nanometer, conversion factor is $\frac{1\text{ nm}}{1.0\times {10}^{-9}\text{ m }}$.