(a)
![5.45 \cdot 10^(14) Hz](https://img.qammunity.org/2020/formulas/physics/high-school/qv9hq0qhzdt5s4e3svnit4eugnwpjiy2if.png)
The relationship between frequency and wavelength of an electromagnetic wave is given by
![c=f \lambda](https://img.qammunity.org/2020/formulas/physics/high-school/pujf2slcux6tuz1i64aby6yqgsc32v3r6t.png)
where
is the speed of light
is the frequency
is the wavelength
In this problem, we are considering light with wavelength of
![\lambda=5.50 \cdot 10^(-7) m](https://img.qammunity.org/2020/formulas/physics/high-school/1ezybmcmxs9f8on0jypdu1jz3cupqocol1.png)
Substituting into the equation and re-arranging it, we can find the corresponding frequency:
![f=(c)/(\lambda)=(3.00 \cdot 10^8 m/s)/(5.50 \cdot 10^(-7) m)=5.45 \cdot 10^(14) Hz](https://img.qammunity.org/2020/formulas/physics/high-school/f9h98gx10ph18doylvjnl55572oz1baqmw.png)
(b)
![1.83\cdot 10^(-15) s](https://img.qammunity.org/2020/formulas/physics/high-school/1dmv6gmpp3swpkijq85lcxed6saz2r6q29.png)
The period of a wave is equal to the reciprocal of the frequency:
![T=(1)/(f)](https://img.qammunity.org/2020/formulas/physics/high-school/c0tafhq7h957t0zih0dt8m9ps0g7o54pyg.png)
And using
as we found in the previous part, we can find the period of this wave:
![T=(1)/(5.45 \cdot 10^(14) Hz)=1.83\cdot 10^(-15) s](https://img.qammunity.org/2020/formulas/physics/high-school/mxwclwx7zpkesvok4q3wi8plg20pms1abs.png)