Question

1. use planks constant, 6.626×10^(-34) j·s to determine how much energy is carried by a photon that has a frequency of 5.71×10^14 hz.

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by the formula $E = hf$, where $E$ is the energy of the photon, $h$ is Planck's constant, and $f$ is the frequency of the photon.

Step2: Substitute given values

We are given that $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$ and $f=5.71\times 10^{14}\text{ Hz}$. Substitute these values into the formula: $E=(6.626\times 10^{-34}\text{ J}\cdot\text{s})\times(5.71\times 10^{14}\text{ Hz})$.

Step3: Perform the multiplication

Using the rules of exponents for multiplication ($a^m\times a^n=a^{m + n}$), we have $E=(6.626\times5.71)\times10^{-34 + 14}\text{ J}$.
$6.626\times5.71 = 37.83446$, and $-34 + 14=-20$. So $E = 3.783446\times 10^{-19}\text{ J}$.

Answer:

$3.78\times 10^{-19}\text{ J}$ (rounded to three significant - figures)