Question

Find the energy of a photon with a wavelength of \(4000\,\text{\AA}\). (Take \( h = 6.63 \times 10^{-34}\,\text{Js} \), \( c = 3 \times 10^8\,\text{m/s} \)).

• A. \(4.97 \times 10^{-19}\,\text{J}\)
• B. \(3.31 \times 10^{-19}\,\text{J}\)
• C. \(6.63 \times 10^{-19}\,\text{J}\)
• D. \(2.48 \times 10^{-19}\,\text{J}\)

Hint:

Always convert wavelength into meters before using \(E = \frac{hc}{\lambda}\).

Answer 1

The Correct Option is A

Concept: Energy of a photon is given by \[ E = \frac{hc}{\lambda} \]

Step 1: Convert wavelength to meters. \[ 1\text{\AA} = 10^{-10}\text{ m} \] \[ \lambda = 4000 \times 10^{-10} \] \[ \lambda = 4 \times 10^{-7}\text{ m} \]

Step 2: Substitute the values. \[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}} \]

Step 3: Simplify. \[ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} \] \[ E = 4.97 \times 10^{-19}\,\text{J} \]