If a source of power 4 kW produces 10^20 photons/second, the radiation belong to a part of the spect

Q: Why do we multiply the number of photons by the energy of one photon?

Power is defined as the total energy delivered per second. If you have $n$ photons arriving every second, and each one carries an energy $E$, the total energy arriving per second is simply $n \times E$. It's like counting the total money you get if $n$ people each give you $E$ dollars every second!

Q: I got confused with the calculation. How do you handle the powers of 10?

When multiplying or dividing numbers in scientific notation, handle the coefficients and the powers of 10 separately. For $10^{20} \times 10^{-34}$, you add the exponents: $20 + (-34) = -14$. Then, when you divide by $10^{-14}$, it moves to the numerator as $10^{14}$. So, $10^3 \times 10^{14} = 10^{17}$.

Q: Do I need to memorize the exact frequencies of the EM spectrum?

You don't need to memorize every single boundary to the decimal point, but you **must** know the order of magnitude. Knowing that visible light is around $10^{14} \text{ Hz}$, X-rays are around $10^{17} \text{ Hz}$, and radio waves are around $10^6 \text{ Hz}$ is crucial for JEE.

Q: What if I calculated the wavelength instead of frequency?

That works perfectly too! You can use $P = n \cdot \frac{hc}{\lambda}$. You would find $\lambda \approx 5 \text{ nm}$. Knowing that X-rays typically range from $0.01 \text{ nm}$ to $10 \text{ nm}$, you would arrive at the exact same conclusion.

Q: Why did the reference solution use $6.023 \times 10^{-34}$?

That is a typographical error in the source material. The author accidentally mixed up Planck's constant ($6.626 \times 10^{-34}$) with the digits of Avogadro's number ($6.022 \times 10^{23}$). Always use the standard value $h \approx 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$.

Question

If a source of power $4 	ext{ kW}$ produces $10^{20}$ photons/second, the radiation belong to a part of the spectrum called

Options:

(A) X-rays

(B) ultraviolet rays

(C) microwaves

(D) $\gamma$-rays

Sigma Percentile · Accepted Answer

### The Power of Photons

Imagine a powerful light source, like a massive searchlight, beaming energy into the void. In classical physics, we think of this energy as a continuous wave. But quantum mechanics tells us a different story: this beam is actually a torrential rain of tiny energy packets called **photons**.

In this problem, we are given a source with a macroscopic power of $4 	ext{ kW}$. That's $4000 	ext{ Joules}$ of energy being pumped out every single second! We are also told that this energy is carried by $10^{20}$ photons per second. Our mission is to find out what kind of light this is.

### Bridging the Macroscopic and Microscopic

To solve this, we need a bridge between the macroscopic world of "Power" and the microscopic world of "Photons".

What exactly is power? **Power** is simply the total energy emitted per unit time. 
If one photon carries an energy $E$, and the source emits $n$ photons every second, then the total energy emitted per second (which is the power $P$) must be the number of photons multiplied by the energy of each photon.

This gives us our master equation:
$$P = n \cdot E$$

We also know from Planck's quantum theory that the energy of a single photon is directly proportional to its frequency $f$. 
$$E = hf$$

Combining these, we get:
$$P = n \cdot hf$$

### Crunching the Numbers

Now, let's plug in the numbers. We have:
$P = 4 	imes 10^3 	ext{ W}$
$n = 10^{20} 	ext{ s}^{-1}$
$h \approx 6.63 	imes 10^{-34} 	ext{ J}\cdot	ext{s}$

Substituting these into our equation:
$$4 	imes 10^3 = 10^{20} \cdot (6.63 	imes 10^{-34}) \cdot f$$

Let's isolate the frequency $f$:
$$f = \frac{4 	imes 10^3}{10^{20} \cdot 6.63 	imes 10^{-34}}$$

$$f = \frac{4 	imes 10^3}{6.63 	imes 10^{-14}}$$

$$f \approx 0.603 	imes 10^{17} 	ext{ Hz}$$

$$f \approx 6.03 	imes 10^{16} 	ext{ Hz}$$

### Decoding the Spectrum

We have found the frequency of the radiation, but the question asks us to identify the *type* of radiation. This requires a mental map of the **Electromagnetic Spectrum**.

Let's recall the typical frequency ranges:
- **Visible Light:** Around $4 	imes 10^{14}$ to $7.5 	imes 10^{14} 	ext{ Hz}$
- **Ultraviolet (UV):** $7.5 	imes 10^{14}$ to $3 	imes 10^{16} 	ext{ Hz}$
- **X-rays:** $3 	imes 10^{16}$ to $3 	imes 10^{19} 	ext{ Hz}$
- **Gamma Rays:** Greater than $3 	imes 10^{19} 	ext{ Hz}$

Our calculated frequency is $6.03 	imes 10^{16} 	ext{ Hz}$. This falls perfectly into the lower end of the **X-ray** region!

Therefore, the source is emitting X-rays. This is a beautiful example of how knowing the macroscopic power and the microscopic particle count allows us to completely identify the nature of the electromagnetic wave.

Animated Solution for Physics - Dual Nature of Matter and Radiation: If a source of power $4 kW$ produces $1 0^{20}$ photons/second, the radiation belong to a part of the spectrum called

Select Answer:

Visualized Solution

$P = 4 kW, n = 1 0^{20} s^{- 1}$

$P = n \cdot h f$

$4 \times 1 0^{3} = 1 0^{20} \cdot h \cdot f$

$f \approx 6 \times 1 0^{16} Hz$

$EM Spectrum$

$Conclusion$

The Sigma Insight: Photon Theory of Light

The Power of Photons

Bridging the Macroscopic and Microscopic

Crunching the Numbers

Decoding the Spectrum

Similar Questions

X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from

The X-ray beam coming from an X-ray tube will be

A $2 mW$ laser operates at a wavelength of $500 nm$ . The number of photons that will be emitted per second is [Given, Planck's constant $h = 6.6 \times 1 0^{- 34} Js$ , speed of light $c = 3.0 \times 1 0^{8} m/s$ ]

An X-ray tube is operated at $1.24$ million volt. The shortest wavelength of the produced photon will be

Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is

A potential difference of $20 kV$ is applied across an X-ray tube. The minimum wavelength of X-rays generated is ....... $\overset{˚}{A}$ .

The shortest wavelength of X-rays emitted from an X-ray tube depends on

Photon of frequency $ν$ has a momentum associated with it. If $c$ is the velocity of light, the momentum is

Four physical quantities are listed in Column I. Their values are listed in Column II in a random order

List-I

List-II

The wavelength of an X-ray beam is $10 \overset{˚}{A}$ . The mass of a fictitious particle having the same energy as that of the X-ray photons is $\frac{x}{3} h kg$ . The value of $x$ is ............ . ( $h = Planck’s constant$ )