Question

18 what is the energy of a photon of red light that has a frequency of 4.48 x 10^14 hz? * (0.5 points) 2.97 x 10^-19 j 1.48 x 10^-48 j 6.76 x 10^47 j 1.34 x 10^23 j 6.70 x 10^-7 j 1.49 x 10^6 j

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by $E = h
u$, where $h$ is Planck's constant ($h=6.63\times 10^{-34}\ J\cdot s$) and $
u$ is the frequency of the photon.

Step2: Substitute values

We are given $
u = 4.48\times 10^{14}\ Hz$. Substitute $h = 6.63\times 10^{-34}\ J\cdot s$ and $
u=4.48\times 10^{14}\ Hz$ into the formula $E = h
u$.
$E=(6.63\times 10^{-34}\ J\cdot s)\times(4.48\times 10^{14}\ Hz)$

Step3: Perform multiplication

Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $E=(6.63\times4.48)\times10^{-34 + 14}\ J$.
$6.63\times4.48 = 29.6924$, and $-34+14=-20$. So $E = 29.6924\times 10^{-20}\ J$.
In scientific - notation, $E = 2.96924\times 10^{-19}\ J\approx2.97\times 10^{-19}\ J$.

Answer:

$2.97\times 10^{-19}\ J$ (corresponding to the first option)