Question

the scattering of sunlight by the mixture of gas molecules in the air results in the blue color of the sky. calculate the energy (in joules) of a single photon of blue light with a frequency of 7.50×10¹⁴ hz. (h = 6.626×10⁻³⁴ j·s) answer: j

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by $E = hf$, where $E$ is energy, $h$ is Planck's constant ($h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$) and $f$ is frequency.

Step2: Substitute given values

We are given $f = 7.50\times 10^{14}\text{ Hz}$ and $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$. Substitute into the formula: $E=(6.626\times 10^{-34}\text{ J}\cdot\text{s})\times(7.50\times 10^{14}\text{ Hz})$.

Step3: Perform multiplication

Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $E=(6.626\times7.50)\times10^{-34 + 14}\text{ J}$.
$6.626\times7.50 = 49.695$, and $-34+14=-20$. So $E = 49.695\times10^{-20}\text{ J}$.

Step4: Scientific - notation adjustment

Rewrite $49.695\times10^{-20}\text{ J}$ in proper scientific - notation. $49.695\times10^{-20}\text{ J}=4.9695\times 10^{-19}\text{ J}$.

Answer:

$4.97\times 10^{-19}\text{ J}$ (rounded to three significant figures)