Question

What is the wavelength associated with a photon of a light with the energy is 3.6e-19j

Answer 1

To find the wavelength associated with a photon given its energy, you can use the equation that relates energy and wavelength:

\[
E = \frac{hc}{\lambda}
\]

Where:
- \( E \) is the energy of the photon (in joules),
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light in vacuum (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength (in meters).

Rearranging the equation to solve for \( \lambda \):

\[
\lambda = \frac{hc}{E}
\]

Now plug in the values:

\[
h = 6.626 \times 10^{-34} \, \text{Js}
\]
\[
c = 3.00 \times 10^8 \, \text{m/s}
\]
\[
E = 3.6 \times 10^{-19} \, \text{J}
\]

Now, substituting the values:

\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^8 \, \text{m/s})}{3.6 \times 10^{-19} \, \text{J}}
\]

Calculating the numerator:

\[
6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Js m/s}
\]

Now, divide by the energy:

\[
\lambda = \frac{1.9878 \times 10^{-25}}{3.6 \times 10^{-19}} \approx 5.52 \times 10^{-7} \, \text{m}
\]

Converting to nanometers (1 m = \( 10^9 \) nm):

\[
\lambda \approx 5.52 \times 10^{-7} \times 10^9 \, \text{nm} \approx 552 \, \text{nm}
\]

So, the wavelength associated with a photon of energy \( 3.6 \times 10^{-19} \) J is approximately **552 nm**.