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(D) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For

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$6.25$

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(D) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a proton,$\lambda_p = \frac{h}{\sqrt{2m_p q_p V_p}}$.For a doubly ionized $^8_4 Be$ nucleus,the mass is $m_{Be} = 8m_p$ and the charge is $q_{Be} = 2q_p$. Thus,$\lambda_{Be} = \frac{h}{\sqrt{2(8m_p)(2q_p)V_{Be}}}$.Given $\lambda_p = \lambda_{Be}$,we equate the denominators:$2m_p q_p V_p = 2(8m_p)(2q_p)V_{Be}$.$m_p q_p V_p = 16 m_p q_p V_{Be}$.$V_p = 16 V_{Be}$.Given $V_p = 100 \ V$,we have $100 = 16 V_{Be}$.$V_{Be} = \frac{100}{16} = 6.25 \ V$.

$A$ proton is accelerated through a potential difference of $100 \ V$. To have the same de Broglie wavelength,what potential difference must be applied across a doubly ionized $_4^8 Be$ nucleus (in $V$)?

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