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

Question

for electromagnetic waves traveling in a vacuum, we can substitute $\frac{c}{\lambda}=f$ for frequency and calculate energy in terms of wavelength.
$e = hf = \frac{hc}{\lambda}$
plancks constant is $6.626×10^{-34}$ joule - seconds (j·s). the speed of light in a vacuum is $2.998×10^{8}$ 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 energy for infrared wave

Use the formula \(E=\frac{hc}{\lambda}\).
Substitute \(h = 6.626\times10^{-34}\space J\cdot s\), \(c = 2.998\times 10^{8}\space m/s\), \(\lambda=3.74\times 10^{-4}\space m\)

$$ LATEXBLOCK0 $$

Step2: Calculate energy for visible light

Substitute \(h = 6.626\times10^{-34}\space J\cdot s\), \(c = 2.998\times 10^{8}\space m/s\), \(\lambda = 4.92\times 10^{-7}\space m\)

$$ LATEXBLOCK1 $$

Step3: Calculate energy for X - ray

Substitute \(h = 6.626\times10^{-34}\space J\cdot s\), \(c = 2.998\times 10^{8}\space m/s\), \(\lambda=1.08\times 10^{-10}\space m\)

$$ LATEXBLOCK2 $$

Answer:

Electromagnetic waveWavelength (m)Energy (J)
Visible light\(4.92\times 10^{-7}\)\(4.04\times 10^{-19}\)
X - ray\(1.08\times 10^{-10}\)\(1.84\times 10^{-16}\)