Question

When light of wavelength '$\lambda$' is incident on a photosensitive surface, photons of power 'P' are emitted. The number of photon 'n' emitted in time 't' is [h = Planck's constant, c = velocity of light in vacuum]

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

Hint:

Total energy is power multiplied by time ($E = P \cdot t$). To get the number of packets, simply divide total energy by single packet energy: $\frac{Pt}{\left(\frac{hc}{\lambda}\right)} = \frac{P\lambda t}{hc}$. Dimensional analysis can also rule out wrong choices immediately!

Answer 1

The Correct Option is C

Step 1: Energy of one photon.
A single photon carries energy $E=\frac{hc}{\lambda}$.

Step 2: Total energy of $n$ photons.
For $n$ photons the energy is $\frac{nhc}{\lambda}$.

Step 3: Bring in power and time.
Power times time equals energy, so $Pt=\frac{nhc}{\lambda}$.

Step 4: Solve for $n$.
\[ n=\frac{P\lambda t}{hc} \] \[ \boxed{\frac{P\lambda t}{hc}} \]