Step 1: Understanding the Concept:
The de Broglie hypothesis explains wave-particle duality, showing that moving particles exhibit wave-like characteristics. The wavelength ($\lambda$) associated with a matter particle depends inversely on its momentum. When comparing different particles, we can rewrite this momentum value in terms of kinetic energy to simplify the comparison.
Step 2: Key Formula or Approach:
The baseline formula for de Broglie wavelength is:
$$ \lambda = \frac{h}{p} $$
We can express momentum ($p$) in terms of mass ($m$) and kinetic energy ($K$) using the classical relationship:
$$ K = \frac{p^2}{2m} \implies p = \sqrt{2mK} $$
Substituting this momentum equation back into the wavelength formula gives:
$$ \lambda = \frac{h}{\sqrt{2mK}} $$
Step 3: Detailed Explanation:
We need to find the ratio of the de Broglie wavelength of an electron ($\lambda_e$) to that of a proton ($\lambda_p$). We are given that both particles share the same kinetic energy ($K_e = K_p = K$).
Let's write out the individual wavelength formulas for both particles:
- For the electron: $\lambda_e = \frac{h}{\sqrt{2m_eK}}$
- For the proton: $\lambda_p = \frac{h}{\sqrt{2m_pK}}$
Now, divide the electron's wavelength equation by the proton's equation to find their ratio:
$$ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_eK}}}{\frac{h}{\sqrt{2m_pK}}} $$
Since Planck's constant ($h$), the constant $2$, and the shared kinetic energy value ($K$) appear in both terms, they cancel out completely:
$$ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_pK}}{\sqrt{2m_eK}} = \sqrt{\frac{m_p}{m_e}} $$
This derived ratio matches option (A).
Step 4: Final Answer:
The ratio of the de Broglie wavelengths is $\sqrt{\frac{m_p}{m_e}}$.