A proton and an alpha-particle are accelerate | vedclass

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Question

(C) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V_0$ is given by $\lambda = \frac{h}{\sqrt{2

Vedclass · Accepted Answer

$2\sqrt{2} : 1$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V_0$ is given by $\lambda = \frac{h}{\sqrt{2mqV_0}}$.Since $h$ and $V_0$ are constant for both particles,the wavelength is inversely proportional to the square root of the product of mass and charge: $\lambda \propto \frac{1}{\sqrt{mq}}$.For a proton,$m_p = m$ and $q_p = e$. For an $\alpha$-particle,$m_{\alpha} = 4m$ and $q_{\alpha} = 2e$.The ratio of the wavelengths is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}q_{\alpha}}{m_pq_p}} = \sqrt{\frac{4m 	imes 2e}{m 	imes e}} = \sqrt{8} = 2\sqrt{2}$.Thus,the ratio is $2\sqrt{2} : 1$.

$A$ proton and an $\alpha$-particle are accelerated through a potential difference of $100\, V$. The ratio of the wavelength associated with the proton to that associated with an $\alpha$-particle is

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