Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

for electromagnetic waves traveling in a vacuum, we can substitute \\( …

Question

for electromagnetic waves traveling in a vacuum, we can substitute \\( \frac { c } { \lambda } = f \\) for frequency and calculate energy in terms of wavelength.

\\( e = h f = \frac { h c } { \lambda } \\)

plancks constant is \\( 6.626 \times 10 ^ { - 34 } \\) joule-seconds (j-s). the speed of light in a vacuum is \\( 2.998 \times 10 ^ { 8 } \\) meters per second.

complete the table to determine the energy carried by each electromagnetic wave traveling in a vacuum.

write your answers in scientific notation rounded to two decimal places.

Explanation:

Step1: Calculate the energy of the infrared wave

Use the formula \(E=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\), \(c=2.998\times 10^{8}\space m/s\), and \(\lambda = 2.55\times10^{-3}\space m\)

$$ LATEXBLOCK0 $$

Step2: Calculate the energy of the visible - light wave

Use the formula \(E=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\), \(c=2.998\times 10^{8}\space m/s\), and \(\lambda = 6.22\times10^{-7}\space m\)

$$ LATEXBLOCK1 $$

Step3: Calculate the energy of the X - ray wave

Use the formula \(E=\frac{hc}{\lambda}\), where \(h = 6.626\times10^{-34}\space J\cdot s\), \(c=2.998\times 10^{8}\space m/s\), and \(\lambda = 5.61\times10^{-10}\space m\)

$$ LATEXBLOCK2 $$

Answer:

Infrared wave: \(7.79\times 10^{-23}\space J\)
Visible light: \(3.20\times 10^{-19}\space J\)
X - ray: \(3.54\times 10^{-16}\space J\)