Question
An $\alpha$-particle and a proton are accelerated through the same potential difference. Find the ratio of their de Broglie wavelength.
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Answer
Finding the ratio of de Broglie wavelength $\bigg(\frac{\lambda{\alpha}}{\lambda{p}}\bigg)$ $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}$ $\therefore\frac{{\lambda}\alpha}{{\lambda}p}=\frac{h}{\sqrt{2m_{\alpha}q_{\alpha}V}}\times\frac{\sqrt{2m_{p}q_p{V}}}{h}$ $\frac{{\lambda}_{\alpha}}{{\lambda}_p}=\frac{\sqrt{m_p{q_p}}}{\sqrt{m_{\alpha}{q_{\alpha}}}}$
$=\frac{\sqrt{m_p{q_p}}}{{\sqrt{4m_p^{\text{ }\text{ }2}{q_p}}}}$
$=\frac{1}{{2}\sqrt{2}}$
$\lambda_{\alpha}:\lambda_p=1:2\sqrt{2}$Need a full question paper?
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