Question

Calculate the energy of a photon of a light with the frequency 6.5e-14s

Answer 1

To calculate the energy of a photon, you can use the formula:

\[
E = h \cdot f
\]

where:
- \( E \) is the energy of the photon (in joules),
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \),
- \( f \) is the frequency of the light (in hertz).

Given that the frequency \( f = 6.5 \times 10^{-14} \, \text{s}^{-1} \) (which is actually expressed in hertz since \( \text{s}^{-1} \) is equivalent to Hz), we can substitute the values into the equation:

\[
E = (6.626 \times 10^{-34} \, \text{Js}) \cdot (6.5 \times 10^{-14} \, \text{s}^{-1})
\]

Now, performing the multiplication:

\[
E = 6.626 \times 6.5 \times 10^{-34} \times 10^{-14}
\]

Calculating \( 6.626 \times 6.5 \):

\[
6.626 \times 6.5 \approx 43.059
\]

Now combining the powers of ten:

\[
E \approx 43.059 \times 10^{-48}
\]

This can be expressed in a more compact scientific notation:

\[
E \approx 4.3059 \times 10^{-47} \, \text{J}
\]

So the energy of the photon with a frequency of \( 6.5 \times 10^{-14} \, \text{s}^{-1} \) is approximately:

\[
E \approx 4.31 \times 10^{-47} \, \text{J}
\]