Question

what is the approximate frequency of a photon having an energy ( 5\times10^{-24} j )? (( h = 6.6\times10^{-34} jcdot s ))
( 1\times10^{-10} hz )
( 3\times10^{-58} hz )
( 8\times10^{9} hz )
( 3\times10^{-57} hz )

Explanation:

Step1: Use the formula \(E = h

u\)
We know that the energy of a photon \(E\) is given by \(E = h
u\), where \(h\) is Planck's constant (\(h = 6.6\times10^{-34}\space J\cdot s\)) and \(
u\) is the frequency of the photon. We need to solve for \(
u\), so we can rewrite the formula as \(
u=\frac{E}{h}\).

Step2: Substitute the given values

Given \(E = 5\times 10^{-24}\space J\) and \(h = 6.6\times10^{-34}\space J\cdot s\). Then \(
u=\frac{5\times 10^{-24}}{6.6\times10^{-34}}\).
Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\), for the powers of 10: \(\frac{10^{-24}}{10^{-34}}=10^{-24+ 34}=10^{10}\). And \(\frac{5}{6.6}\approx0.76\). So \(
u\approx0.76\times10^{10}=7.6\times10^{9}\space Hz\approx8\times10^{9}\space Hz\)

Answer:

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