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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.For a proton,$m_p = m$ and $q_p = e$. Thus,$\lambda_p = \fra

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$3$

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.For a proton,$m_p = m$ and $q_p = e$. Thus,$\lambda_p = \frac{h}{\sqrt{2meV}}$.For an $\alpha$-particle,$m_\alpha = 4m$ and $q_\alpha = 2e$. Thus,$\lambda_\alpha = \frac{h}{\sqrt{2(4m)(2e)V}} = \frac{h}{\sqrt{16meV}} = \frac{1}{4} \frac{h}{\sqrt{meV}}$.Comparing the two,$\lambda_\alpha = \frac{1}{4\sqrt{2}} \lambda_p$ is incorrect; let us re-evaluate: $\lambda_\alpha = \frac{h}{\sqrt{2(4m)(2e)V}} = \frac{h}{4\sqrt{meV}}$.Therefore,$\frac{\lambda_p}{\lambda_\alpha} = \frac{h / \sqrt{2meV}}{h / \sqrt{16meV}} = \frac{\sqrt{16}}{\sqrt{2}} = \sqrt{8} \approx 2.828$.The nearest integer to $2.828$ is $3$.

An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 \ V$. After this,their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_{p}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_\alpha}$,to the nearest integer,is

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