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(A) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$.Since the potential diffe

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$\lambda_p = 2\sqrt{2} \lambda_{\alpha}$

Vedclass · Answer

(A) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$.Since the potential difference $V$ and Planck's constant $h$ are the same for both particles,we have $\lambda \propto \frac{1}{\sqrt{mq}}$.For a proton,$q_p = e$ and $m_p = m_p$.For an $\alpha$-particle,$q_{\alpha} = 2e$ and $m_{\alpha} = 4m_p$.Taking the ratio: $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{(4m_p)(2e)}{(m_p)(e)}} = \sqrt{8} = 2\sqrt{2}$.Therefore,$\lambda_p = 2\sqrt{2} \lambda_{\alpha}$.

$A$ proton and an $\alpha$-particle are accelerated using the same potential difference. How are the de-Broglie wavelengths $\lambda_p$ and $\lambda_{\alpha}$ related to each other?

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