Question

1. use plancks constant, (h = 6.63\times10^{-34} jcdot s) to determine how much energy is carried by a photon that has a frequency of (5.71\times10^{14} hz).

• (1.14\times10^{9} j)
• (3.78\times10^{-19} j)
• (3.70\times10^{21} j)
• (8.88\times10^{47} j)

Explanation:

Step1: Recall the energy - frequency formula

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

Step2: Substitute the given values

We are given $h = 6.63\times10^{-34}\ J\cdot s$ and $f = 5.71\times 10^{14}\ Hz$. Then $E=(6.63\times 10^{-34}\ J\cdot s)\times(5.71\times 10^{14}\ Hz)$.
Using the rule of exponents for multiplication $a^m\times a^n=a^{m + n}$, we have $E=(6.63\times5.71)\times10^{-34 + 14}\ J$.
$6.63\times5.71 = 37.8573$ and $-34 + 14=-20$. So $E = 37.8573\times10^{-20}\ J=3.78573\times 10^{-19}\ J\approx3.78\times 10^{-19}\ J$.

Answer:

$3.78\times 10^{-19}\ J$ (corresponding to the second - option in the multiple - choice list you provided, assuming the options are in order as shown in the image)