Question

When light of wavelength \( \lambda \) is incident on photosensitive surface, photons of power \( P \) are emitted. The number of photons \( n \) emitted in \( t \) second is

• A. \( \frac{hc}{P \lambda t} \)
• B. \( \frac{P \lambda t}{hc} \)
• C. \( \frac{P \lambda}{htc} \)
• D. \( \frac{hP}{\lambda tc} \)

Hint:

When calculating the number of photons, use the relation between power, energy of a photon, and the time interval. The energy of a photon is inversely proportional to its wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
The energy of a photon is given by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the velocity of light, and \( \lambda \) is the wavelength. The power emitted by the photons is \( P = nE \), where \( n \) is the number of photons emitted per second.
Step 2: Deriving the formula.
The total energy emitted in time \( t \) is \( P \times t \). The number of photons \( n \) is given by the ratio of the total energy to the energy per photon: \[ n = \frac{P \times t}{E} = \frac{P \times t}{\frac{hc}{\lambda}} = \frac{P \lambda t}{hc} \] Step 3: Conclusion.
Thus, the correct answer is (B) \( \frac{P \lambda t}{hc} \).