Question

the energy of a photon is related to frequency by e = hf, where h = 6.63 × 10 - 34 joule seconds. burning a pound of coal liberates about 10,000,000 joules of heat energy. knowing that the frequency of visible light is 6 × 1014 hz, how many visible photons are needed to have an energy equivalent to the heat energy liberated from burning a pound of coal?
a. 0.522e - 23
b. 2.22e20
c. 2.52e25
d. 25.2e - 2

Explanation:

Step1: Calculate the energy of one photon

The formula for the energy of a photon is \(E = hf\), where \(h = 6.63\times10^{-34}\space J\cdot s\) and \(f=6\times 10^{14}\space Hz\).
Substitute the values into the formula:
\(E=(6.63\times 10^{-34})\times(6\times 10^{14})\)
Using the rule of exponents \(a^m\times a^n=a^{m + n}\), we have \(E = 6.63\times6\times10^{-34 + 14}\)
\(E=39.78\times10^{-20}=3.978\times 10^{-19}\space J\)

Step2: Calculate the number of photons

We know that the total energy \(E_{total}=10000000\space J\)
The number of photons \(n=\frac{E_{total}}{E}\)
Substitute \(E_{total}=10000000\space J\) and \(E = 3.978\times 10^{-19}\space J\) into the formula:
\(n=\frac{10000000}{3.978\times 10^{-19}}\)
Using the rule \(\frac{a}{b\times10^{-n}}=\frac{a}{b}\times10^{n}\), we get \(n=\frac{1\times10^{7}}{3.978\times 10^{-19}}\)
\(n=\frac{1}{3.978}\times10^{7+ 19}\)
\(n\approx0.2514\times10^{26}=2.514\times10^{25}\approx2.52\times10^{25}\)

Answer:

C. \(2.52E25\)