Question

what is the approximate frequency of a photon having an energy 5 × 10-24 j? (h = 6.6 × 10-34 js)
○ 3 × 10-38 hz
○ 1 × 10-10 hz
○ 8 × 109 hz
○ 3 × 10-57 hz

Explanation:

Step1: Use the formula \(E = hf\)

We know that \(E = 5\times10^{-24}\text{ J}\) and \(h=6.6\times 10^{-34}\text{ Js}\). Rearranging the formula for \(f\) gives \(f=\frac{E}{h}\).

Step2: Substitute the values

Substitute \(E = 5\times10^{-24}\) and \(h = 6.6\times10^{-34}\) into \(f=\frac{E}{h}\). So \(f=\frac{5\times 10^{-24}}{6.6\times10^{-34}}\).
Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\), we have \(f=\frac{5}{6.6}\times10^{-24+ 34}\).
\(\frac{5}{6.6}\approx0.76\), and \(10^{-24 + 34}=10^{10}\). So \(f\approx0.76\times10^{10}=7.6\times10^{9}\approx8\times10^{9}\text{ Hz}\)

Answer:

\(8\times 10^{9}\text{ Hz}\) (the third option)