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The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume $$m_p = m_e \times 1849$$)
The de-Broglie wavelength is given by:
$$\lambda = \frac{h}{\sqrt{2mK}}$$
where $$K$$ is the kinetic energy.
For same kinetic energy:
$$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}}$$
Given $$m_p = 1849 \times m_e$$:
$$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{1849 \times m_e}} = \frac{1}{\sqrt{1849}} = \frac{1}{43}$$
The ratio $$\lambda_p : \lambda_e = 1 : 43$$.
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