Question

31 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 1.90 x 10^29 j 2.11 x 10^12 j 9.54 x 10^53 j 5.19 x 10^-23 j 1.05 x 10^-54 j

Explanation:

Step1: Recall the energy - frequency formula

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

Step2: Substitute the given values

We are given that $
u = 6.32\times 10^{20}\ Hz$. Substituting $h = 6.63\times 10^{-34}\ J\cdot s$ and $
u=6.32\times 10^{20}\ Hz$ into the formula $E = h
u$, we get $E=(6.63\times 10^{-34}\ J\cdot s)\times(6.32\times 10^{20}\ Hz)$.

Step3: Perform the multiplication

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

Answer:

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