Question

6 calculate the energy possessed by a single photon whose frequency is 6.32 x 10^20 hz. * (0.5 points) 4.19 x 10^-13 j 5.19 x 10^-23 j 1.90 x 10^29 j 1.05 x 10^-54 j 9.54 x 10^53 j 2.11 x 10^12 j

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by $E = hf$, where $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ (Planck's constant) and $f$ is the frequency of the photon.

Step2: Substitute values

We are given $f = 6.32\times 10^{20}\text{ Hz}$. Substitute $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ and $f = 6.32\times 10^{20}\text{ Hz}$ into the formula:
$E=(6.63\times 10^{-34}\text{ J}\cdot\text{s})\times(6.32\times 10^{20}\text{ Hz})$

Step3: Perform multiplication

Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $E=(6.63\times6.32)\times10^{-34 + 20}\text{ J}$.
$6.63\times6.32 = 41.9016$ and $-34+20=-14$. So $E = 41.9016\times10^{-14}\text{ J}$.
In scientific - notation, $E = 4.19016\times 10^{-13}\text{ J}\approx4.19\times 10^{-13}\text{ J}$.

Answer:

$4.19\times 10^{-13}\text{ J}$