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  4. A proton when accelerated through a potential difference of V, has a de-Broglie wavelength λ associated with it. If an α -particle is to have the same de-Broglie wavelength λ, it must be accelerated through a potential difference of

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Q. A proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$ -particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

EAMCETEAMCET 2012

Solution:

$\lambda_{p}=\lambda_{\alpha}$
$(m q V)_{p}=(m q V)_{\alpha}$
Potential difference
$V_{\alpha}=\frac{V}{8} [ \because m_{\alpha} =4 m_{p}
q_{\alpha} =2 q_{p}]$