Question
One of the spectral lines of caesium has a wavelength of 456 nm . Calculate the frequency of this line $\left(\mathrm{c}=3.0 \times 10^8 \mathrm{~ms}^{-1}\right)$.
✓
Answer
$\lambda=456\text{nm}=456\times10^{-9}\text{m},$
$\text{v}=\ ?,\text{c}=3.0\times10^8\text{ms}^{-1}$
$\text{v}=\frac{\text{c}}{\lambda}=\frac{3.0\times10^8\text{ms}^{-1}}{456\times10^{-9}\text{m}}$
$=6.578\times10^{14}\text{Hz}$
$\text{v}=\ ?,\text{c}=3.0\times10^8\text{ms}^{-1}$
$\text{v}=\frac{\text{c}}{\lambda}=\frac{3.0\times10^8\text{ms}^{-1}}{456\times10^{-9}\text{m}}$
$=6.578\times10^{14}\text{Hz}$
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