The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1

Q: Why do we convert nanometers to meters?

In physics, maintaining dimensional consistency is critical. Planck's constant ($h$) is given in Joule-seconds, and the speed of light ($c$) is in meters per second. To ensure the units cancel out correctly and leave us with Joules for energy, the wavelength must be in meters.

Q: Why did we approximate $19.89$ to $20$ during the calculation?

In competitive exams like JEE, examiners often design numerical values to cancel out elegantly. Approximating $19.89$ to $20$ introduces a negligible error (about $0.5\%$) but drastically simplifies the subsequent division, saving precious time without affecting the final nearest integer answer.

Q: What is the exact difference between power and energy in this context?

Think of energy as a volume of water, and power as the rate at which water flows from a tap. Energy ($E$) is the capacity to do work (measured in Joules), while Power ($P$) is how fast that energy is delivered (measured in Joules per second, or Watts).

Q: Can we use the shortcut formula $E = 1240 \text{ eV nm} / \lambda$?

Yes, you absolutely can! However, that formula gives the energy in Electron-Volts (eV). Since the power is given in Watts (Joules/second), you would have to convert the total energy back into Joules at the end ($1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$). In this specific case, sticking to standard SI units from the start is slightly more direct.

Q: Why do we multiply by $0.1$ at the very end instead of the beginning?

You can actually do it either way! You could calculate the total energy emitted in $0.1$ seconds first ($E_{total} = P \times t = 10^{-4} \text{ J}$), and then divide that total energy by the energy of a single photon. Both logical paths are perfectly valid and will yield the exact same result.

Question

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power $1	ext{ mW}$ and wavelength of $1000	ext{ nm}$, in $0.1	ext{ second}$ is $x 	imes 10^{13}$. The value of $x$ is ......... . (Nearest integer)
($h= 6.63 	imes 10^{-34}	ext{ Js}, c = 3.00 	imes 10^8	ext{ ms}^{-1}$)

Sigma Percentile · Accepted Answer

### The Quantum Nature of Light

Imagine a tiny infrared laser pointer. When you turn it on, it emits a continuous beam of light. However, classical physics tells us only half the story.

According to **Planck's Quantum Theory**, this beam is not a continuous wave. Instead, it is composed of discrete, indivisible packets of energy called **photons**.

In this problem, we are tasked with counting the exact number of these microscopic energy packets emitted by a $1	ext{ mW}$ laser in a brief time window of $0.1	ext{ seconds}$.

### Decoding the Energy of a Single Photon

To count the total number of photons, we must first determine the energy carried by a single photon.

The energy $E$ of a photon is inversely proportional to its wavelength $\lambda$, given by the fundamental equation:
$$E = \frac{hc}{\lambda}$$

Here, $h$ is **Planck's constant** ($6.63 	imes 10^{-34}	ext{ Js}$) and $c$ is the **speed of light** ($3.00 	imes 10^8	ext{ ms}^{-1}$).

The wavelength is given as $1000	ext{ nm}$. Before substituting, we must convert this into standard SI units (meters) to maintain dimensional consistency. 
$$1000	ext{ nm} = 1000 	imes 10^{-9}	ext{ m} = 10^{-6}	ext{ m}$$

Now, we substitute these values into our energy equation:
$$E = \frac{(6.63 	imes 10^{-34}) 	imes (3.00 	imes 10^8)}{10^{-6}}$$

Calculating the numerator gives $19.89 	imes 10^{-26}$. Dividing by $10^{-6}$ yields:
$$E = 19.89 	imes 10^{-20}	ext{ J}$$

For the sake of computational elegance, we can approximate this to $20 	imes 10^{-20}	ext{ J}$. This is the energy of just one single photon.

### The Power-Photon Relationship

Now that we know the energy of an individual photon, we need to relate it to the macroscopic power of the laser.

**Power** is defined as the total energy emitted per unit time (Joules per second).

If the laser emits $n$ photons every second, the total power $P$ is simply the number of photons multiplied by the energy of one photon:
$$P = n 	imes E$$

We are given that the power of the laser is $1	ext{ mW}$, which is $10^{-3}	ext{ W}$ (or $10^{-3}	ext{ Joules per second}$).

Substituting our known values into the power equation:
$$10^{-3} = n 	imes (20 	imes 10^{-20})$$

Solving for $n$, the rate of photon emission:
$$n = \frac{10^{-3}}{20 	imes 10^{-20}} = 0.05 	imes 10^{17} = 0.5 	imes 10^{16}	ext{ photons/second}$$

### Calculating the Total Emission

We have found that the laser fires $0.5 	imes 10^{16}$ photons every single second.

However, the problem specifically asks for the number of photons emitted in a time window of $t = 0.1	ext{ seconds}$.

To find the total number of photons $N$, we multiply the emission rate $n$ by the time $t$:
$$N = n 	imes t$$
$$N = (0.5 	imes 10^{16}) 	imes 0.1$$
$$N = 0.5 	imes 10^{15}	ext{ photons}$$

### The Final Formatting

We have the total number of photons, but the question requires the answer in a very specific format: $x 	imes 10^{13}$.

We need to mathematically manipulate our result to match this format without changing its actual value. 
$$0.5 	imes 10^{15} = 50 	imes 10^{13}$$

By comparing this with the given expression $x 	imes 10^{13}$, it becomes immediately clear that the value of $x$ is exactly **$50$**.

This problem beautifully demonstrates how macroscopic properties like power are directly built upon the microscopic, quantum properties of individual photons!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Emission}

\text{Energy of a Single Photon}

\text{Substituting Values}

\text{Calculating Photon Energy}

\text{Power and Emission Rate}

\text{Calculating Photons per Second}

\text{Total Photons in Given Time}

\text{Finding } x

\text{Food for Thought}

The Sigma Insight: Wave Particle Duality

The Quantum Nature of Light

Decoding the Energy of a Single Photon

The Power-Photon Relationship

Calculating the Total Emission

The Final Formatting

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