Question

What is the ration of Debroglie wavemength of proton and neutron if kinetic energy is same for both

Hint:

For particles with the same kinetic energy, their de Broglie wavelengths are inversely proportional to the square root of their masses ($\lambda \propto 1/\sqrt{m}$). Since proton and neutron have roughly the same mass, their wavelengths are roughly equal.

Answer 1

Step 1: Understanding the Concept:
The de Broglie wavelength ($\lambda$) of a particle is related to its momentum ($p$) by the equation $\lambda = h/p$. Momentum can be expressed in terms of kinetic energy ($K$) and mass ($m$).

Step 2: Key Formula or Approach:
The relationship between momentum and kinetic energy is $K = p^2 / 2m$, which gives $p = \sqrt{2mK}$.
Substituting this into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

Step 3: Detailed Explanation:
For a proton with mass $m_p$ and kinetic energy $K$, its de Broglie wavelength is:
\[ \lambda_p = \frac{h}{\sqrt{2m_p K}} \]
For a neutron with mass $m_n$ and the same kinetic energy $K$, its de Broglie wavelength is:
\[ \lambda_n = \frac{h}{\sqrt{2m_n K}} \]
Taking the ratio of their wavelengths:
\[ \frac{\lambda_p}{\lambda_n} = \frac{\frac{h}{\sqrt{2m_p K}}}{\frac{h}{\sqrt{2m_n K}}} = \sqrt{\frac{2m_n K}{2m_p K}} = \sqrt{\frac{m_n}{m_p}} \]
The mass of a neutron ($m_n \approx 1.675 \times 10^{-27}$ kg) is very slightly larger than the mass of a proton ($m_p \approx 1.672 \times 10^{-27}$ kg). For most approximate calculations in such problems, their masses are considered nearly equal ($m_n \approx m_p$).
Therefore, $\frac{\lambda_p}{\lambda_n} \approx \sqrt{1} = 1$.

Step 4: Final Answer:
The ratio is approximately $1:1$.