A particle of mass 9.1 x 10^-31 kg travels in a medium with a speed of 10^6 m/s and a photon of a ra

Q: Why do we use the same formula $\lambda = h/p$ for both a particle and a photon?

The de-Broglie relation $\lambda = h/p$ is a universal principle that applies to all matter and radiation, linking their wave and particle natures.

Q: Do we need the value of Planck's constant to solve this?

No, because we are asked for the ratio of the wavelengths. Planck's constant $h$ appears in the numerator for both and cancels out perfectly.

Q: Can a photon have momentum if it has no rest mass?

Yes! While a photon has zero rest mass, it has energy ($E = hc/\lambda$) and momentum ($p = E/c = h/\lambda$) due to its relativistic nature.

Q: What if the particle was moving at a speed close to the speed of light?

If the particle's speed was comparable to $c$, we would have to use the relativistic momentum formula $p = \gamma mv$ instead of the classical $p = mv$.

Q: Why is the photon's wavelength so much larger than the particle's?

The particle has a much larger momentum ($9.1 \times 10^{-25}$ kg m/s) compared to the photon ($10^{-27}$ kg m/s). Since wavelength is inversely proportional to momentum, the photon has a larger wavelength.

Question

A particle of mass $9.1 	imes 10^{-31} 	ext{ kg}$ travels in a medium with a speed of $10^6 	ext{ m/s}$ and a photon of a radiation of linear momentum $10^{-27} 	ext{ kg-m/s}$ travels in vacuum. The wavelength of photon is ......... times the wavelength of the particle.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two completely different entities traveling through space. On one hand, we have a tiny particle—perhaps an electron—with a mass of $m = 9.1 	imes 10^{-31} 	ext{ kg}$ cruising at a speed of $v = 10^6 	ext{ m/s}$. On the other hand, we have a massless photon of light zipping through a vacuum, carrying a momentum of $p_{ph} = 10^{-27} 	ext{ kg m/s}$.

Our mission is to compare their wave-like natures by finding the ratio of their wavelengths.

### The Master Equation

To bridge the gap between particles and waves, we invoke the brilliant de-Broglie hypothesis. It states that every moving object has an associated wavelength given by:

$$ \lambda = \frac{h}{p} $$

For our massive particle, the momentum is the product of its mass and velocity ($p_{pa} = mv$). So, its de-Broglie wavelength is:

$$ \lambda_{pa} = \frac{h}{mv} $$

For the photon, the wavelength is directly related to its momentum:

$$ \lambda_{ph} = \frac{h}{p_{ph}} $$

### Setting Up the Ratio

We are asked to find how many times the photon's wavelength is compared to the particle's wavelength. Let's set up the ratio:

$$ \frac{\lambda_{ph}}{\lambda_{pa}} = \frac{\frac{h}{p_{ph}}}{\frac{h}{mv}} $$

Notice something beautiful? The Planck's constant $h$ cancels out completely! We don't even need to plug in its messy value. The expression simplifies elegantly to:

$$ \frac{\lambda_{ph}}{\lambda_{pa}} = \frac{mv}{p_{ph}} $$

### Final Calculation

Now, let's substitute the given values into our simplified ratio.

$$ \frac{\lambda_{ph}}{\lambda_{pa}} = \frac{9.1 	imes 10^{-31} 	imes 10^6}{10^{-27}} $$

First, let's combine the powers of 10 in the numerator:

$$ \frac{\lambda_{ph}}{\lambda_{pa}} = \frac{9.1 	imes 10^{-25}}{10^{-27}} $$

Finally, bringing the $10^{-27}$ up to the numerator:

$$ \frac{\lambda_{ph}}{\lambda_{pa}} = 9.1 	imes 10^{-25} 	imes 10^{27} = 9.1 	imes 10^2 $$

$$ \frac{\lambda_{ph}}{\lambda_{pa}} = 910 $$

The photon's wavelength is exactly **910** times the wavelength of the particle. This problem beautifully demonstrates how momentum dictates the wave nature of both matter and light!

Enter Numerical Value:

Visualized Solution

$Visual Anchor$

$Logic Bridge$

$Raw Setup$

$Atomic Compute$

$Final Answer$

$The Way Forward$

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Setup

The Master Equation

Setting Up the Ratio

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Anchor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Logic Bridge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Raw Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Atomic Compute

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward