Question

What is the energy, in eV, of (a) a 450 MHz radio-frequency photon, (b) a visible-light photon with a wavelength of 450 nm, and (c) an x-ray photon with a wavelength of 0.045 nm?

Short answer

a) Energy is equal to $1.86\times {10}^{-6}eV$.

b) Energy is equal to 2.76 eV.

c) Energy is equal to 27.6 keV.

Step-by-step solution

Step 1. Given Information is :a) a 450 MHz radio-frequency photon b) a visible-light photon with a wavelength of 450 nmc) an x-ray photon with a wavelength of 0.045 nm 

We need to find out the energy of the given photons.

Step 2. Part a) Using expression of Energy of a photon.

$$\begin{gathered} E=hf \\ E=(6.63\times {10}^{-34})(450\times {10}^{6}Hz) \\ E=2.98\times {10}^{-25}J \\ E=1.86\times {10}^{-6}eV \end{gathered}$$

Step 3. Part b) Using expression of Energy of a photon. 

$$\begin{gathered} E=hf \\ f=\frac{c}{\lambda } \\ E=h\frac{c}{\lambda } \\ E=\frac{(6.63\times {10}^{-34})(3\times {10}^{8}m/s)}{450\times {10}^{-9}m} \\ E=4.42\times {10}^{-19}J \\ E=2.76eV \end{gathered}$$

Step 4. Part c) Using expression of Energy of a photon. 

$$\begin{gathered} E=\frac{hc}{\lambda } \\ E=\frac{(6.63\times {10}^{-34})(3\times {10}^{8}m/s)}{\left(0.045\times {10}^{-9}m\right)} \\ E=4.42\times {10}^{-15}J \\ E=27.6keV \end{gathered}$$