The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Le

Q: Why do we use different formulas for the wavelength of a photon and a proton?

A photon is a massless particle of light, so its energy is strictly related to its wavelength by $E = \frac{hc}{\lambda}$. A proton is a massive particle, so its wavelength is given by the de-Broglie relation $\lambda = \frac{h}{p}$.

Q: How did we get $p = \sqrt{2mK}$?

Kinetic energy is $K = \frac{1}{2}mv^2$. Multiplying the numerator and denominator by $m$ gives $K = \frac{m^2v^2}{2m} = \frac{p^2}{2m}$. Solving for $p$ gives $p = \sqrt{2mK}$.

Q: Why does Planck's constant cancel out?

Both the de-Broglie wavelength and the photon energy formula fundamentally rely on Planck's constant to relate wave properties to particle properties. When taking the ratio, this universal constant naturally cancels out.

Q: What happens if the energy $E$ is doubled?

Since the ratio is proportional to $E^{1/2}$, doubling the energy will increase the ratio by a factor of $\sqrt{2}$.

Q: Can we apply the photon formula to the proton?

No, the formula $E = \frac{hc}{\lambda}$ is derived from $E = pc$ which is only valid for massless particles traveling at the speed of light. Protons have rest mass and travel slower than light.

Question

The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. The ratio $\frac{\lambda_1}{\lambda_2}$ is proportional to

Options:

(A) $E^0$

(B) $E^{1/2}$

(C) $E^{-1}$

(D) $E^{-2}$

Sigma Percentile · Accepted Answer

This problem presents a fascinating scenario where the energy of a massless photon is perfectly equal to the kinetic energy of a massive proton. Let's denote this shared energy value simply as $E$. This will be the bridge connecting our two different particles.

### Analyzing the Proton

Let's focus our attention on the proton. Since it's a particle with mass, we need to find its de-Broglie wavelength, which we'll call $\lambda_1$. The fundamental relation states that wavelength equals Planck's constant divided by momentum:

$$\lambda_1 = \frac{h}{p}$$

We can relate momentum to kinetic energy using the formula $p = \sqrt{2mK}$. Substituting our given kinetic energy $E$, we get:

$$\lambda_1 = \frac{h}{\sqrt{2mE}}$$

### Analyzing the Photon

Next, let's shift gears and look at the photon. For a photon, the relationship between energy and wavelength is much more direct. The energy $E$ is simply equal to Planck's constant times the speed of light, divided by its wavelength, $\lambda_2$:

$$E = \frac{hc}{\lambda_2}$$

By rearranging this simple equation, we can easily isolate $\lambda_2$:

$$\lambda_2 = \frac{hc}{E}$$

### Finding the Ratio

We have our two wavelengths, so what's next? The problem asks for the ratio of $\lambda_1$ to $\lambda_2$. Let's set up the division. We take our expression for $\lambda_1$ and divide it by the expression for $\lambda_2$. Instead of dividing by a fraction, we can multiply by its reciprocal:

$$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc}$$

Now comes the fun part, simplifying the algebra! Look closely at the expression. Planck's constant, $h$, appears in both the numerator and the denominator, so it cancels out beautifully. We are left with $E$ in the numerator and the square root of $E$ in the denominator:

$$\frac{\lambda_1}{\lambda_2} = \frac{1}{c\sqrt{2m}} \cdot \frac{E}{\sqrt{E}}$$

When we divide $E$ by $E^{1/2}$, we are left with $E^{1/2}$ in the numerator. The remaining terms, $c$ and $\sqrt{2m}$, are just constants:

$$\frac{\lambda_1}{\lambda_2} = \frac{1}{c\sqrt{2m}} \cdot E^{1/2}$$

### The Final Conclusion

Finally, let's look at our simplified ratio. We have a bunch of constants multiplied by $E^{1/2}$. Since we only care about the proportionality with respect to $E$, we can ignore all those constants. This clearly shows us that the ratio of the wavelengths is directly proportional to $E^{1/2}$.

**$$\frac{\lambda_1}{\lambda_2} \propto E^{1/2}$$**

And there we have it, our final answer!

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Visualized Solution

$Given Information$

$Wavelength of Proton$

$Wavelength of Photon$

$Ratio of Wavelengths$

$Simplifying the Ratio$

$Final Proportionality$

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Proton

Analyzing the Photon

Finding the Ratio

The Final Conclusion

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Visualized Solution

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

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

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

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

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

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