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Question 40.5: The Energy Range of Visible Photons We are able to see light......

The Energy Range of Visible Photons

We are able to see light in the wavelength range from about 400 nm to about 700 nm. What is the corresponding range in the energy of photons that we can detect with our eyes? Report your answer to one significant figure in joules and in electron-volts.

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INTERPRET and ANTICIPATE
An electromagnetic wave’s frequency can be found from its wavelength. The photon’s energy is directly proportional to the wave’s frequency.

SOLVE
Find the frequencies using Equation 34.20. (It is okay to keep an extra significant figure in this step.)

\begin{aligned}& \lambda f=c \quad \quad (34.20) \\& f=c / \lambda \\& \text { Red: } f_{\text {red }}=\frac{3.00 \times 10^8 m / s }{700 \times 10^{-9} m }=4.3 \times 10^{14} Hz \\& \text { Violet: } f_{\text {violet }}=\frac{3.00 \times 10^8 m / s }{400 \times 10^{-9} m }=7.5 \times 10^{14} Hz\end{aligned}

Find the corresponding energies from the frequencies.

\begin{aligned}& E=h f \quad \quad (40.8) \\& \text { Red: } E_{ red }=\left(6.626 \times 10^{-34} J \cdot s \right)\left(4.3 \times 10^{14} Hz \right)=3 \times 10^{-19} J \\& E_{ red } \approx 2  eV \\& \text { Violet: } E_{\text {violet }}=\left(6.626 \times 10^{-34} J \cdot s \right)\left(7.5 \times 10^{14} Hz \right)=5 \times 10^{-19} J \\& E_{\text {violet }} \approx 3  eV \\&\end{aligned}

CHECK AND THINK
So we are able to detect photons with energies between approximately 2 and 3 eV.

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