Question

1. what is the energy of light whose wavelength is 4.06×10⁻¹¹ m?
2. a helium laser emits light with a wavelength of 633 nm. what is the frequency of the light?
3. what is the wavelength of x - rays having a frequency of 4.80×10¹⁷ hz?
4. an fm radio station broadcasts at a frequency of 107.9 mhz. what is the wavelength of the radio signal? (hint: first convert megahertz mhz into hertz.)
5. if the limits of human hearing are 20 hz to 20,000 hz, what are the sound wavelengths that are associated with both of these two extremes, assuming the speed of sound is 345 m/s.

frequency = 20 hz: wavelength =
frequency = 20,000 hz: wavelength =

1. rank these parts of the electromagnetic spectrum from lowest energy (1) to highest (7): gamma, infrared, microwave, radio, visible, ultraviolet, x - ray
2. rank these parts of the electromagnetic spectrum from lowest frequency (a) to highest (g): gamma, infrared, microwave, radio, visible, ultraviolet, x - ray
3. rank these parts of the electromagnetic spectrum from shortest wavelength (a) to longest (g): gamma, infrared, microwave, radio, visible, ultraviolet, x - ray
4. circle one: what is the relationship between frequency and wavelength? (direct or inverse)
5. circle one: what is the relationship between frequency and energy? (direct or inverse)

Explanation:

Step1: Recall wave - speed formula

The speed of light $c = \lambda f$, where $c = 3\times10^{8}\ m/s$ (speed of light in vacuum), $\lambda$ is wavelength and $f$ is frequency. Also, the energy of a photon is $E=hf$, where $h = 6.626\times10^{- 34}\ J\cdot s$ (Planck's constant).

Step2: Solve for frequency in problem 10

Given $\lambda=633\ nm = 633\times10^{-9}\ m$. From $c = \lambda f$, we can solve for $f$: $f=\frac{c}{\lambda}=\frac{3\times10^{8}}{633\times10^{-9}}\ Hz\approx4.74\times10^{14}\ Hz$

Step3: Solve for wavelength in problem 11

Given $f = 4.80\times10^{17}\ Hz$. From $c=\lambda f$, we get $\lambda=\frac{c}{f}=\frac{3\times10^{8}}{4.80\times10^{17}}\ m = 6.25\times10^{-10}\ m$

Step4: Solve for wavelength in problem 12

Given $f = 107.9\ MHz=107.9\times10^{6}\ Hz$. From $c=\lambda f$, we have $\lambda=\frac{c}{f}=\frac{3\times10^{8}}{107.9\times10^{6}}\ m\approx2.78\ m$

Step5: Solve for wavelengths in problem 13

For $f_1 = 20\ Hz$, using $c = \lambda f$ (where $c = 345\ m/s$ for sound), $\lambda_1=\frac{c}{f_1}=\frac{345}{20}\ m = 17.25\ m$
For $f_2=20000\ Hz$, $\lambda_2=\frac{c}{f_2}=\frac{345}{20000}\ m = 0.01725\ m\approx0.02\ m$

Step6: Rank by energy in problem 14

The energy of electromagnetic radiation is $E = hf$, and $c=\lambda f$. Lower frequency means lower energy. The order from lowest to highest energy is: Radio (1), Microwave (2), Infrared (3), Visible (4), Ultraviolet (5), X - ray (6), Gamma (7)

Step7: Rank by frequency in problem 15

The order from lowest to highest frequency is: Radio (a), Microwave (b), Infrared (c), Visible (d), Ultraviolet (e), X - ray (f), Gamma (g)

Step8: Rank by wavelength in problem 16

Since $c=\lambda f$, higher frequency means shorter wavelength. The order from shortest to longest wavelength is: Gamma (A), X - ray (B), Ultraviolet (C), Visible (D), Infrared (E), Microwave (F), Radio (G)

Step9: Answer relationship questions 18 and 19

The relationship between frequency $f$ and wavelength $\lambda$ is inverse ($c=\lambda f$). The relationship between frequency $f$ and energy $E$ is direct ($E = hf$)

Answer:

1. $4.74\times10^{14}\ Hz$
2. $6.25\times10^{-10}\ m$
3. $2.78\ m$
4. For $20\ Hz$, wavelength is $17.25\ m$; for $20000\ Hz$, wavelength is $0.02\ m$
5. Radio (1), Microwave (2), Infrared (3), Visible (4), Ultraviolet (5), X - ray (6), Gamma (7)
6. Radio (a), Microwave (b), Infrared (c), Visible (d), Ultraviolet (e), X - ray (f), Gamma (g)
7. Gamma (A), X - ray (B), Ultraviolet (C), Visible (D), Infrared (E), Microwave (F), Radio (G)
8. Inverse
9. Direct