A photon has an energy of 0.30 eV. What is its wavelength in nanometers (nm)? (Use 1 eV = 1.602 × 10^{-19} J, h = 6.626 × 10^{-34} J·s, and c = 3.00 × 10^{8} m/s.)

• A. 1240 nm
• B. 4140 nm
• C. 248 nm
• D. 826 nm

Answer: (B)

Verified step by step guidance

1. Identify the relationship between the energy of a photon and its wavelength. The energy \(E\) of a photon is related to its wavelength \(\lambda\) by the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant and \(c\) is the speed of light.
2. Rearrange the equation to solve for the wavelength \(\lambda\): \(\lambda = \frac{hc}{E}\).
3. Convert the given energy from electronvolts (eV) to joules (J) using the conversion factor \$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}\(. So, calculate \)E\( in joules as \)E = 0.30\ \text{eV} \times 1.602 \times 10^{-19}\ \text{J/eV}$.
4. Substitute the values of \(h = 6.626 \times 10^{-34}\ \text{J}\cdot\text{s}\), \(c = 3.00 \times 10^{8}\ \text{m/s}\), and the converted energy \(E\) into the wavelength formula \(\lambda = \frac{hc}{E}\) to find \(\lambda\) in meters.
5. Convert the wavelength from meters to nanometers by multiplying by \$10^{9}\( (since \)1\ \text{m} = 10^{9}\ \text{nm}$) to get the final wavelength in nanometers.