Question

example calculations 5.1.3 determine the frequency in hz and the wavelength in m of a photon with an energy of 5.204×10^(-19) j. a. $
u=\frac{5.204\times 10^{-19}j}{6.62607\times 10^{-34}jcdot s}=7.854\times 10^{14}hz$ b. $lambda=\frac{(6.62607\times 10^{-34}jcdot s)\times(2.99792\times 10^{8}m/s)}{5.204\times 10^{-19}j}=3.817\times 10^{-7}m$ question 5.1.7 determine the frequency in hz and the wavelength in m of a photon with an energy of 3.205×10^(-19) j. use e - format for scientific notation and enter four digits (1.234×10^(-5)=1.234e - 5) do not enter the unit. a. $
u = 4.837\times 10^{-7}hz$ b. $lambda=6.198\times 10^{-7}m$

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by $E = h
u$, where $E$ is energy, $h = 6.62607\times10^{-34}\text{ Js}$ is Planck's constant and $
u$ is frequency. So, $
u=\frac{E}{h}$.
Substitute $E = 3.205\times 10^{-19}\text{ J}$ and $h = 6.62607\times10^{-34}\text{ Js}$ into the formula:
$
u=\frac{3.205\times 10^{-19}\text{ J}}{6.62607\times10^{-34}\text{ Js}}\approx4.837\times 10^{14}\text{ Hz}$ (in scientific - notation $4.837E14$).

Step2: Recall energy - wavelength formula

The energy of a photon can also be written as $E=\frac{hc}{\lambda}$, where $c = 2.99792\times10^{8}\text{ m/s}$ is the speed of light and $\lambda$ is the wavelength. So, $\lambda=\frac{hc}{E}$.
Substitute $h = 6.62607\times10^{-34}\text{ Js}$, $c = 2.99792\times10^{8}\text{ m/s}$ and $E = 3.205\times 10^{-19}\text{ J}$ into the formula:
$\lambda=\frac{(6.62607\times10^{-34}\text{ Js})\times(2.99792\times10^{8}\text{ m/s})}{3.205\times 10^{-19}\text{ J}}\approx6.198\times 10^{-7}\text{ m}$ (in scientific - notation $6.198E - 7$).

Answer:

a. $4.837E14$
b. $6.198E - 7$