Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respecti

Q: Why is the number of photons directly proportional to the wavelength?

Because the energy of a single photon is inversely proportional to its wavelength ($E = hc/\lambda$). To maintain a constant total power, a source emitting lower-energy photons (larger wavelength) must emit more of them per second.

Q: Does 'number density' here mean photons per unit volume?

In the context of this problem, 'number density' refers to the number of photons emitted per unit time (photon flux). The problem uses it interchangeably with the rate of photon emission to compare the two sources.

Q: Why did the official answer key say 500 instead of 1/500?

The official answer key calculated the ratio of visible light photons to X-ray photons ($n_V / n_X = 500$). However, the question explicitly asks for the ratio of X-rays to visible light, which is mathematically $1/500$. It's a known discrepancy in the original exam's key!

Q: What happens if the power of the X-ray source was doubled?

If the power of the X-ray source was doubled to 400 W, it would need to emit twice as many photons per second. The new ratio $n_X / n_V$ would then become $2/500$ or $1/250$.

Q: Do we need to convert the wavelengths from nanometers to meters to solve this?

Not at all! Since we are taking a ratio ($\lambda_X / \lambda_V$), the units cancel out completely. As long as both wavelengths are in the same units (nanometers), you can plug them in directly.

Question

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

Options:

(A) $\frac{1}{250}$

(B) 500

(C) 250

(D) $\frac{1}{500}$

Sigma Percentile · Accepted Answer

The problem of comparing two light sources—one emitting X-rays and the other visible light—is a classic exploration of the **Dual Nature of Radiation and Matter**. It beautifully bridges the macroscopic world of power (Watts) with the microscopic world of quantum photons. Let's dive into the physics behind this and uncover a fascinating catch in the official answer key!

### Analyzing the Setup

Imagine you are standing in a lab with two distinct light sources. The first source is a high-energy X-ray emitter, shooting out invisible rays with a tiny wavelength of $\lambda_X = 1	ext{ nm}$. The second source is a bright visible light lamp, beaming at a much larger wavelength of $\lambda_V = 500	ext{ nm}$.

Despite their drastic differences in wavelength, both sources are glowing with the exact same macroscopic power: $P = 200	ext{ W}$.

Our mission is to find the ratio of the number of X-ray photons emitted per second to the number of visible light photons emitted per second. To do this, we need to translate "Power" into the language of quantum mechanics.

### The Master Equation

What does power actually mean at the quantum level? Power is defined as the total energy emitted per unit of time. If a source emits $n$ photons every second, and each photon carries an energy $E$, the total power $P$ is simply the product of the two:

$$P = n \cdot E$$

Now, we bring in Max Planck and Albert Einstein. The energy of a single photon is inversely proportional to its wavelength, given by the famous equation:

$$E = \frac{hc}{\lambda}$$

Substituting this into our power equation, we get the master equation for this problem:

$$P = n \cdot \frac{hc}{\lambda}$$

### The Quantum Tug-of-War

Let's rearrange our master equation to solve for $n$, the number of photons emitted per second:

$$n = \frac{P \lambda}{hc}$$

Look closely at this relationship. For both of our light sources, the power $P$ is exactly $200	ext{ W}$. Planck's constant $h$ and the speed of light $c$ are universal constants. Because these three values are identical for both sources, they act as a constant scaling factor.

This reveals a beautiful, direct proportionality:

$$n \propto \lambda$$

This makes perfect physical sense! X-ray photons have a very small wavelength, meaning each individual photon packs a massive punch of energy. To maintain a power of $200	ext{ W}$, you only need a relatively small number of these "heavy-hitters."

Visible light photons, on the other hand, have a much larger wavelength and lower energy. To reach that same $200	ext{ W}$ output, the source must pump out a massive army of these weaker photons.

### Final Calculation and The Catch

The question specifically asks for the ratio of the number density of X-ray photons ($n_X$) to the number density of visible light photons ($n_V$). Using our proportionality, we can set up the ratio:

$$\frac{n_X}{n_V} = \frac{\lambda_X}{\lambda_V}$$

Now, we simply substitute the given wavelengths:

$$\frac{n_X}{n_V} = \frac{1	ext{ nm}}{500	ext{ nm}} = \frac{1}{500}$$

The mathematical and physical ratio is exactly $\frac{1}{500}$.

**A Word of Caution:** If you look at the official JEE answer key, it marks option (b) $500$ as the correct answer. However, $500$ is the ratio of visible photons to X-ray photons ($\frac{n_V}{n_X}$). The question explicitly asked for X-rays to visible light. While we must acknowledge the official key's typo, as physicists, we trust the math. The true ratio is $\frac{1}{500}$!

Select Answer:

Visualized Solution

Given Parameters $P_{X} = P_{V}$

Power in terms of Photons $P = n \cdot E$

Photon Energy $E = \frac{h c}{λ}$

Proportionality $n \propto λ$

Setting up the Ratio $\frac{n _{X}}{n _{V}}$

Final Calculation $\frac{1}{500}$

Conceptual Takeaway

The Sigma Insight: Photon Theory of Light

Analyzing the Setup

The Master Equation

The Quantum Tug-of-War

Final Calculation and The Catch

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Select Answer:

Visualized Solution

Given Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PX​=PV​

Power in terms of Photons .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=n⋅E

Photon Energy .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=λhc​

Proportionality .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n∝λ

Setting up the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nV​nX​​

Final Calculation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5001​