Question

Light of wavelength \(400\ \text{nm}\) falls on a metal surface. The energy of one photon is approximately: \[ \left( h=6.63\times10^{-34}\ \text{J s},\quad c=3\times10^8\ \text{m/s} \right) \]

• A. \(4.97\times10^{-19}\ \text{J}\)
• B. \(2.5\times10^{-19}\ \text{J}\)
• C. \(1.2\times10^{-18}\ \text{J}\)
• D. \(6.63\times10^{-34}\ \text{J}\)

Hint:

Photon energy: \[ E=\frac{hc}{\lambda} \] Smaller wavelength means: \[ \text{higher photon energy} \]

Answer 1

The Correct Option is A

Energy of one photon is given by: \[ E=\frac{hc}{\lambda} \] where:
• \(h=6.63\times10^{-34}\ \text{J s}\)
• \(c=3\times10^8\ \text{m/s}\)
• \(\lambda=400\ \text{nm}=400\times10^{-9}\ \text{m}\)
Step 1: Substitute the values \[ E= \frac{ (6.63\times10^{-34}) (3\times10^8) }{ 400\times10^{-9} } \]
Step 2: Simplify \[ E= \frac{ 19.89\times10^{-26} }{ 4\times10^{-7} } \] \[ E= 4.97\times10^{-19}\ \text{J} \] Thus: \[ \boxed{ E=4.97\times10^{-19}\ \text{J} } \] Option analysis:
• Option (A): Correct
• Option (B): Incorrect
• Option (C): Incorrect
• Option (D): Incorrect Therefore: \[ \boxed{\text{(A)}} \]