What is the ratio of the de Broglie wavelengt | vedclass

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(A) The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda =

Vedclass · Accepted Answer

$1/\sqrt{2}$

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(A) The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Since both particles are accelerated through the same potential difference $V$ and have the same charge $q = e$,the wavelength is inversely proportional to the square root of the mass: $\lambda \propto \frac{1}{\sqrt{m}}$.Let $m_p$ be the mass of the proton and $m_d$ be the mass of the deuteron. We know that $m_d \approx 2m_p$.The ratio of the wavelengths is $\frac{\lambda_d}{\lambda_p} = \sqrt{\frac{m_p}{m_d}} = \sqrt{\frac{m_p}{2m_p}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

What is the ratio of the de Broglie wavelengths of a deuteron and a proton accelerated through the same potential difference?

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