"An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are $\lambda$$\alpha$ and $\lambda$p respectively. The ratio ${{{{\lambda _p}} \over {{\lambda _\alpha }}}}$ is :"

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Accepted Answer

"We know,

$qv = {{{p^2}} \over {2m}}$

$\Rightarrow p = \sqrt {2mqv}$

$\therefore$ $\lambda = {h \over {\sqrt {2mqv} }}$

$\therefore$ $${{{\lambda _p}} \over {{\lambda _\alpha }}} = {{\sqrt {2{m_\alpha }{q_\alpha }v} } \over {\sqrt {2{m_p}{q_p}v} }}$$

$= {{\sqrt {2(4m)(2e)} } \over {\sqrt {2me} }}$

$= \sqrt 8$

$= 2\sqrt 2$"

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are $\lambda$_$\alpha$ and $\lambda$_p respectively. The ratio ${{{{\lambda _p}} \over {{\lambda _\alpha }}}}$ is :

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An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are $\lambda$$\alpha$ and $\lambda$p respectively. The ratio ${{{{\lambda _p}} \over {{\lambda _\alpha }}}}$ is :

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An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200V. After this, their de Broglie wavelengths are $\lambda$_$\alpha$ and $\lambda$_p respectively. The ratio ${{{{\lambda _p}} \over {{\lambda _\alpha }}}}$ is :