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Step-by-Step Solution
Step 1: Identify the Relevant Concept
When a charged particle (electron or proton) is accelerated through the same potential difference $V$, it gains kinetic energy $K = qV$ (where $q$ is the magnitude of the charge on the particle). The de-Broglie wavelength of any particle is given by $ \lambda = \frac{h}{p} $, where $p$ is the momentum of the particle and $h$ is Planck’s constant.
Step 2: Express the Momentum of Each Particle
For a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$, the kinetic energy is
$ K = qV = \frac{1}{2}mv^2 $.
Hence,
$ \frac{1}{2} m v^2 = qV \implies v = \sqrt{\frac{2 q V}{m}}. $
Consequently, the momentum
$ p = m v = m \sqrt{\frac{2 q V}{m}} = \sqrt{2 m q V}. $
Step 3: Write the de-Broglie Wavelength for Each Particle
Substituting $ p = \sqrt{2 m q V} $ into the de-Broglie relation $ \lambda = \frac{h}{p} $, we obtain
$ \lambda = \frac{h}{\sqrt{2 m q V}}. $
Step 4: Form the Ratio of Wavelengths (Electron to Proton)
For the electron ($m_e$, $q = e$) and proton ($m_p$, $q = e$) accelerated through the same potential difference $V$, their de-Broglie wavelengths are:
$ \lambda_e = \frac{h}{\sqrt{2\,m_e\,e\,V}} $
$ \lambda_p = \frac{h}{\sqrt{2\,m_p\,e\,V}} $
The ratio $ \frac{\lambda_e}{\lambda_p} $ is:
$$
\frac{\lambda_e}{\lambda_p}
= \frac{\frac{h}{\sqrt{2\,m_e\,e\,V}}}{\frac{h}{\sqrt{2\,m_p\,e\,V}}}
= \sqrt{\frac{m_p}{m_e}}.
$$
Step 5: State the Conclusion
The ratio of the de-Broglie wavelength of the electron to that of the proton, when both are accelerated through the same potential difference, is
$ \sqrt{\frac{m_p}{m_e}}. $
