The ratio of de Broglie wavelength of a deute | vedclass

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$.For a deuteron $(d)$,the mass $

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$2$

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$.For a deuteron $(d)$,the mass $m_d = 2m_p$ and kinetic energy $KE_d = E$.For an alpha particle $(\alpha)$,the mass $m_{\alpha} = 4m_p$ and kinetic energy $KE_{\alpha} = 2E$.The ratio of wavelengths is $\frac{\lambda_d}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha} \cdot KE_{\alpha}}{m_d \cdot KE_d}}$.Substituting the values: $\frac{\lambda_d}{\lambda_{\alpha}} = \sqrt{\frac{4m_p \cdot 2E}{2m_p \cdot E}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$.Thus,the ratio is $2:1$,which means $n = 2$.

The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$ is $n:1$. The value of $n$ is . . . . . . .
(Assume mass of proton = mass of neutron)

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The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$ is $n:1$. The value of $n$ is . . . . . . .(Assume mass of proton = mass of neutron)