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(B) The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda =

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$\frac{\lambda_0}{2\sqrt{2}}$

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(B) The de Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Since $h$ and $V$ are constant,$\lambda \propto \frac{1}{\sqrt{mq}}$.For a proton,$m_p = m$ and $q_p = e$. Thus,$\lambda_0 \propto \frac{1}{\sqrt{m_p e}}$.For an alpha particle,$m_{\alpha} = 4m_p$ and $q_{\alpha} = 2e$.Let the wavelength of the alpha particle be $\lambda_{\alpha}$. Then $\lambda_{\alpha} \propto \frac{1}{\sqrt{m_{\alpha} q_{\alpha}}} = \frac{1}{\sqrt{(4m_p)(2e)}} = \frac{1}{\sqrt{8m_p e}}$.Taking the ratio: $\frac{\lambda_{\alpha}}{\lambda_0} = \sqrt{\frac{m_p e}{8m_p e}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$.Therefore,$\lambda_{\alpha} = \frac{\lambda_0}{2\sqrt{2}}$.

The de Broglie wavelength of a proton accelerated by a potential difference of $100 \ V$ is $\lambda_0$. If an alpha particle is accelerated by the same potential difference,its de Broglie wavelength will be:

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