To calculate the energy of a photon, we use the equation:

$E = \frac{hc}{\lambda}$

Where:

• $E$ is the energy of the photon,
• $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{J·s}$),
• $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$),
• $\lambda$ is the wavelength of the photon.

Given: $\lambda = 0.05 \, \text{nm} = 0.05 \times 10^{-9} \, \text{m}$,

Substitute the values:

$E = \frac{(6.626 \times 10^{-34}) (3.00 \times 10^8)}{0.05 \times 10^{-9}}$

$E = \frac{1.9878 \times 10^{-25}}{0.05 \times 10^{-9}}$

$E = 3.9756 \times 10^{-15} \, \text{J}$

To convert this energy to electron volts (eV), divide by the charge of an electron ($1.602 \times 10^{-19} \, \text{C}$):

$E = \frac{3.9756 \times 10^{-15}}{1.602 \times 10^{-19}} \, \text{eV}$

$E \approx 24,820 \, \text{eV} = 24.82 \, \text{keV}$

Final Answer:

The energy of the photon is approximately 24.82 keV.

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Do you have any questions or need further clarification?

Related Questions:

1. How does photon energy vary with wavelength?
2. What is the energy of a photon with a wavelength of 0.1 nm?
3. Can photon energy exceed 1 MeV, and how?
4. What is the relationship between photon energy and frequency?
5. How is Planck's constant used in quantum mechanics?

Tip:

Photon energy is inversely proportional to wavelength, so shorter wavelengths correspond to higher energies.