Question

Determine the energy of a photon in eV if its wavelength is \( 4000\ \text{\AA} \).

• A. \( 3.1 \, \text{eV} \)
• B. \( 2.5 \, \text{eV} \)
• C. \( 1.55 \, \text{eV} \)
• D. \( 4.2 \, \text{eV} \)

Hint:

Remember the shortcut: \( E(\text{eV}) = \frac{12400}{\lambda(\text{\AA})} \) for quick calculations in photon energy problems.

Answer 1

The Correct Option is A

Concept: Energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] To calculate energy in electron volts (eV), we use the convenient relation: \[ E(\text{eV}) = \frac{12400}{\lambda(\text{\AA})} \]
Step 1: {Write the given wavelength.} \[ \lambda = 4000\ \text{\AA} \]
Step 2: {Use the standard formula.} \[ E = \frac{12400}{4000} \]
Step 3: {Compute the value.} \[ E = 3.1 \, \text{eV} \]
Step 4: {Verification using SI units (optional).} \[ E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4000 \times 10^{-10}} \] \[ E \approx 4.97 \times 10^{-19} \, \text{J} \] Convert to eV: \[ 1\,\text{eV} = 1.6 \times 10^{-19} \, \text{J} \] \[ E \approx \frac{4.97}{1.6} \approx 3.1 \, \text{eV} \]