Question

1. the wavelength of a wave is measured from.

a. crest to crest
b. crest to trough
c. origin to crest
d. origin to trough

1. what is the wavelength of a radio wave having a frequency of 3.75 x 107 hz?

a..08 m
b. 0.8 m
c. 8.0 m
d. 80 m

1. which of the following power sources makes use of the photoelectric effect?

a. solar power cells
b. nuclear power plants
c. alkaline batteries
d. diesel engines

1. what is the energy of a single photon of light from hydrogens 656 nm emission?

use h = 6.626 x 10-34 j-s and c = 3.00 x 108 m/s.
a. 1.30 x 10-31 j
b. 1.45 x 10-48 j
c. 3.30 x 1018 j
d. 3.03 x 10-19 j

1. which of the following electromagnetic waves has photons of the highest energy?

a. microwaves
b. infrared
c. x - rays
d. ultraviolet

1. what is the frequency of yellow light, which has a wavelength of 5.56 x 10-7 m?

a. 1.85 x 1015 hz
b. 1.85 x 1015 m/s
c. 5.40 x 1014 hz
d. 5.40 x 1014 m/s

1. the part of the electromagnetic spectrum that humans can see is the _.

a. visible light
b. infrared spectrum
c. ultraviolet spectrum
d. a, b, and c

1. a wavelength of 500 nm is associated with the _ portion of the electromagnetic spectrum.

a. visible
b. infrared
c. ultraviolet
d. microwave

Explanation:

1. Wavelength Definition

Wavelength is the distance between two consecutive identical points on a wave (e.g., crest - to - crest or trough - to - trough).

2. Wavelength - Frequency Relationship

The formula for the speed of a wave \(c = \lambda f\), where \(c = 3\times10^{8}\ m/s\) (speed of light in vacuum for electromagnetic waves), \(f = 3.75\times 10^{7}\ Hz\). Then \(\lambda=\frac{c}{f}=\frac{3\times 10^{8}}{3.75\times 10^{7}} = 8\ m\).

3. Photoelectric Effect Application

The photoelectric effect is the emission of electrons when light shines on a material. Solar power cells use this effect to convert light energy into electrical energy.

4. Photon Energy Calculation

The formula for photon energy \(E = h\frac{c}{\lambda}\). Given \(\lambda=656\ nm=656\times 10^{-9}\ m\), \(h = 6.626\times 10^{-34}\ J\cdot s\), \(c = 3\times 10^{8}\ m/s\). Then \(E=6.626\times 10^{-34}\times\frac{3\times 10^{8}}{656\times 10^{-9}}\approx3.03\times 10^{-19}\ J\).

5. Energy of Electromagnetic Waves

The energy of a photon \(E = hf\) (or \(E = h\frac{c}{\lambda}\)). Since \(E\propto\frac{1}{\lambda}\) (for \(c\) and \(h\) constant), \(x -\) rays have the shortest wavelength among the given options (\(microwaves(\lambda\sim10^{-2}\ m - 1\ m)\), \(infrared(\lambda\sim700\ nm - 1\ mm)\), \(ultraviolet(\lambda\sim10\ nm - 400\ nm)\), \(x - rays(\lambda\sim0.01\ nm - 10\ nm)\)), so \(x -\) rays have the highest energy.

6. Frequency - Wavelength Relationship

Using \(c=\lambda f\), \(f=\frac{c}{\lambda}\), with \(c = 3\times 10^{8}\ m/s\), \(\lambda=5.56\times 10^{-7}\ m\). Then \(f=\frac{3\times 10^{8}}{5.56\times 10^{-7}}\approx5.40\times 10^{14}\ Hz\).

7. Visible Light

The part of the electromagnetic spectrum that humans can see is called visible light (\(\lambda\sim400\ nm - 700\ nm\)). Infrared has longer wavelengths and ultraviolet has shorter wavelengths than what the human eye can detect.

8. Electromagnetic Spectrum Region

\(500\ nm\) is in the range of visible light (\(400\ nm - 700\ nm\)). Infrared has \(\lambda>700\ nm\), ultraviolet has \(\lambda < 400\ nm\), and microwaves have much longer wavelengths (\(\lambda\sim10^{-2}\ m - 1\ m\)).

Answer:

1. a. crest to crest
2. c. \(8.0\ m\)
3. a. solar power cells
4. d. \(3.03\times 10^{-19}\ J\)
5. c. \(x -\) rays
6. c. \(5.40\times 10^{14}\ Hz\)
7. a. visible light
8. a. visible