Question

what is the frequency of a radar photon with energy 6.80×10-25 j? (h = 6.63×10-34 j.s)

Explanation:

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

We know the energy formula for a photon is \(E = hf\), where \(E\) is energy, \(h\) is Planck's constant (\(h = 6.63\times10^{-34}\text{ Js}\)), and \(f\) is frequency. We need to solve for \(f\), so we can rewrite the formula as \(f=\frac{E}{h}\).

Step2: Substitute the given values

Given \(E = 6.80\times 10^{-25}\text{ J}\) and \(h = 6.63\times10^{-34}\text{ Js}\), substitute into \(f=\frac{E}{h}\). So \(f=\frac{6.80\times 10^{-25}}{6.63\times10^{-34}}\).

Step3: Calculate the value

When we divide \(6.80\times 10^{-25}\) by \(6.63\times10^{-34}\), we use the rule of exponents \(\frac{a\times10^{m}}{b\times10^{n}}=\frac{a}{b}\times10^{m - n}\). Here \(a = 6.80\), \(b=6.63\), \(m=- 25\), \(n=-34\). So \(f=\frac{6.80}{6.63}\times10^{-25+34}\). \(\frac{6.80}{6.63}\approx1.03\) and \(10^{-25 + 34}=10^{9}\). So \(f\approx1.03\times10^{9}\text{ Hz}\)

Answer:

\(1.03\times10^{9}\text{ Hz}\)