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

Step 1: Understanding the Question:
The objective is to calculate the energy associated with a specific wavelength of light.
The topic is Modern Physics (Dual Nature of Radiation).
Step 2: Key Formula or Approach:
Use Planck's equation for photon energy:
\[ E = \frac{hc}{\lambda} \]
Ensure all units are in the SI system (meters, Joules).
Step 3: Detailed Explanation:
Given:
Wavelength \( \lambda = 4000\,\text{\AA} = 4000 \times 10^{-10}\,\text{m} = 4 \times 10^{-7}\,\text{m} \).
Constants: \( h = 6.63 \times 10^{-34}\,\text{Js} \), \( c = 3 \times 10^8\,\text{m/s} \).
Substitute values into the formula:
\[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{4 \times 10^{-7}} \]
\[ E = \frac{19.89 \times 10^{-26}}{4 \times 10^{-7}} \]
\[ E = 4.9725 \times 10^{-19}\,\text{J} \]
Step 4: Final Answer:
Rounding off, the energy is \( 4.97 \times 10^{-19}\,\text{J} \).