Question

on each question below, show all work, and put a box around your final answer. once you have found your answer, color the appropriate sections with the right color on the front of the page. 1 is the relationship between wavelength and frequency direct or inverse? 2 what is the wavelength of a wave having a frequency of 3.76 x 10^14 hz? 3 what is the frequency of a wave carrying 8.35 x 10^-18 j of energy? 4 what is the wavelength of a 4.34 x 10^15 hz wave? 5 what is the energy of a 3.12 x 10^18 hz wave? 6 what is the wavelength (in meters) of the electromagnetic carrier wave transmitted by the sports fan radio station at a frequency of 6.40 x 10^4 hz? 7 what is the wavelength of a 1.528 x 10^-13 j wave? 8 calculate the energy of electromagnetic radiation with wavelength of 3.48 x 10^-10 m. 9 calculate the energy of light that has a frequency of 8.71 x 10^5 hz. 10 what is the energy of a 9.33 x 10^-3 m wave?

Explanation:

Step1: Recall wave - related formulas

The speed of light formula is $c = \lambda
u$, where $c = 3.00\times10^{8}\text{ m/s}$ is the speed of light in a vacuum, $\lambda$ is the wavelength and $
u$ is the frequency. The energy - frequency formula is $E=h
u$, where $h = 6.63\times10^{- 34}\text{ J}\cdot\text{s}$ is Planck's constant.

Step2: Solve question 1

The relationship between wavelength ($\lambda$) and frequency ($
u$) is inverse. This is because $c=\lambda
u$, and $c$ is a constant in a vacuum. So, as $\lambda$ increases, $
u$ decreases and vice - versa.

Step3: Solve question 2

Given $
u = 3.76\times10^{14}\text{ Hz}$, using $\lambda=\frac{c}{
u}$, we substitute $c = 3.00\times10^{8}\text{ m/s}$ and $
u = 3.76\times10^{14}\text{ Hz}$ into the formula.
$\lambda=\frac{3.00\times10^{8}\text{ m/s}}{3.76\times10^{14}\text{ Hz}}\approx7.98\times10^{-7}\text{ m}$

Step4: Solve question 3

Given $E = 8.35\times10^{-18}\text{ J}$, using $
u=\frac{E}{h}$, we substitute $E = 8.35\times10^{-18}\text{ J}$ and $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ into the formula.
$
u=\frac{8.35\times10^{-18}\text{ J}}{6.63\times10^{-34}\text{ J}\cdot\text{s}}\approx1.26\times10^{16}\text{ Hz}$

Step5: Solve question 4

Given $
u = 4.34\times10^{15}\text{ Hz}$, using $\lambda=\frac{c}{
u}$, we substitute $c = 3.00\times10^{8}\text{ m/s}$ and $
u = 4.34\times10^{15}\text{ Hz}$ into the formula.
$\lambda=\frac{3.00\times10^{8}\text{ m/s}}{4.34\times10^{15}\text{ Hz}}\approx6.91\times10^{-8}\text{ m}$

Step6: Solve question 5

Given $
u = 3.12\times10^{18}\text{ Hz}$, using $E = h
u$, we substitute $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ and $
u = 3.12\times10^{18}\text{ Hz}$ into the formula.
$E=(6.63\times10^{-34}\text{ J}\cdot\text{s})\times(3.12\times10^{18}\text{ Hz})\approx2.07\times10^{-15}\text{ J}$

Step7: Solve question 6

Given $
u = 6.40\times10^{4}\text{ Hz}$, using $\lambda=\frac{c}{
u}$, we substitute $c = 3.00\times10^{8}\text{ m/s}$ and $
u = 6.40\times10^{4}\text{ Hz}$ into the formula.
$\lambda=\frac{3.00\times10^{8}\text{ m/s}}{6.40\times10^{4}\text{ Hz}}\approx4.69\times10^{3}\text{ m}$

Step8: Solve question 7

First, find the frequency using $
u=\frac{E}{h}$. Given $E = 1.528\times10^{-13}\text{ J}$ and $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$, $
u=\frac{1.528\times10^{-13}\text{ J}}{6.63\times10^{-34}\text{ J}\cdot\text{s}}\approx2.30\times10^{20}\text{ Hz}$. Then, using $\lambda=\frac{c}{
u}$, we substitute $c = 3.00\times10^{8}\text{ m/s}$ and $
u\approx2.30\times10^{20}\text{ Hz}$ into the formula.
$\lambda=\frac{3.00\times10^{8}\text{ m/s}}{2.30\times10^{20}\text{ Hz}}\approx1.30\times10^{-12}\text{ m}$

Step9: Solve question 8

First, find the frequency using $
u=\frac{c}{\lambda}$. Given $\lambda = 3.48\times10^{-10}\text{ m}$ and $c = 3.00\times10^{8}\text{ m/s}$, $
u=\frac{3.00\times10^{8}\text{ m/s}}{3.48\times10^{-10}\text{ m}}\approx8.62\times10^{17}\text{ Hz}$. Then, using $E = h
u$, we substitute $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ and $
u\approx8.62\times10^{17}\text{ Hz}$ into the formula.
$E=(6.63\times10^{-34}\text{ J}\cdot\text{s})\times(8.62\times10^{17}\text{ Hz})\approx5.72\times10^{-16}\text{ J}$

Step10: Solve question 10

First, find the frequency using $
u=\frac{c}{\lambda}$. Given $\lambda = 9.33\times10^{-3}\text{ m}$ and $c = 3.00\times10^{8}\text{ m/s}$, $
u=\frac{3.00\times10^{8}\text{ m/s}}{9.33\times10^{-3}\text{ m}}\approx3.22\times10^{10}\text{ Hz}$. Then, using $E = h
u$, we substitute $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ and $
u\approx3.22\times10^{10}\text{ Hz}$ into the formula.
$E=(6.63\times10^{-34}\text{ J}\cdot\text{s})\times(3.22\times10^{10}\text{ Hz})\approx2.14\times10^{-23}\text{ J}$

Answer:

1. Inverse
2. $7.98\times10^{-7}\text{ m}$
3. $1.26\times10^{16}\text{ Hz}$
4. $6.91\times10^{-8}\text{ m}$
5. $2.07\times10^{-15}\text{ J}$
6. $4.69\times10^{3}\text{ m}$
7. $1.30\times10^{-12}\text{ m}$
8. $5.72\times10^{-16}\text{ J}$
9. $5.77\times10^{-28}\text{ J}$ (using $E = h

u$ with $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ and $
u = 8.71\times10^{5}\text{ Hz}$)

1. $2.14\times10^{-23}\text{ J}$