Proton $(\mathrm{P})$ and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $\mathrm{m}_{\mathrm{p}}=1849 \mathrm{~m}_{\mathrm{e}}$ ):
Solution
<p>The de Broglie wavelength of a particle is given by the formula:</p>
<p>$\lambda = \frac{h}{p}$</p>
<p>where $h$ is Planck's constant and $p$ is the momentum of the particle.</p>
<p>If the de Broglie wavelengths of the proton and electron are the same, then:</p>
<p>$\frac{h}{p_p} = \frac{h}{p_e}$</p>
<p>where $p_p$ and $p_e$ are the momenta of the proton and electron, respectively.</p>
<p>Solving this equation for the ratio of their momenta gives:</p>
<p>$\frac{p_p}{p_e} = 1$</p>
<p>So, the ratio of their momenta is 1:1</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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