Question

Kinetic energy of a proton is equal to energy '$E$' of a photon. Let '$\lambda_1$' be the de-Broglie wavelength of proton and '$\lambda_2$' be the wavelength of photon. If $\frac{\lambda_1}{\lambda_2} \propto E^n$, then the value of '$n$' is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $2$
• D. $4$

Hint:

To solve scaling problems like this quickly, focus only on the energy variable power tracker: $\lambda_1 \propto \frac{1}{\sqrt{E}} \rightarrow E^{-1/2}$, and $\lambda_2 \propto \frac{1}{E} \rightarrow E^{-1}$. Taking their ratio gives: $$\frac{\lambda_1}{\lambda_2} \propto \frac{E^{-1/2}}{E^{-1}} = E^{-1/2 - (-1)} = E^{1/2}$$ This confirms $n = 1/2$ within seconds without writing out any full physical constants!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Determine the exponent n in a wavelength-energy scaling relation by tracking only the energy power dependence.

Step 2: Key Formula or Approach:
Wavelength λ₁ ∝ 1/√E = E^(-1/2). Wavelength λ₂ ∝ 1/E = E^(-1). Form the ratio λ₁/λ₂ and simplify the exponents algebraically.

Step 3: Detailed Explanation:
Computing the ratio: λ₁/λ₂ ∝ E^(-1/2)/E^(-1) = E^(-1/2 - (-1)) = E^(-1/2 + 1) = E^(1/2). The resulting exponent n = 1/2 emerges purely from the difference of the individual energy exponents. Focusing exclusively on the power tracker avoids writing out Planck's constant, speed of light, or any other physical constants, condensing the derivation to seconds.

Step 4: Final Answer:
The exponent n equals 1/2.