The kinetic energies of an electron,alpha-par | vedclass

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$.For an electron: $m_e \approx \frac{m_p}{1840}$,$K_e = 4K$. Thus,$\lambda_e

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$\lambda_\alpha < \lambda_p < \lambda_e$

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mK}}$.For an electron: $m_e \approx \frac{m_p}{1840}$,$K_e = 4K$. Thus,$\lambda_e = \frac{h}{\sqrt{2(m_p/1840)(4K)}} = \frac{h}{\sqrt{2m_pK/230}}$.For a proton: $m_p = m$,$K_p = K$. Thus,$\lambda_p = \frac{h}{\sqrt{2mK}}$.For an $\alpha$-particle: $m_\alpha = 4m_p$,$K_\alpha = 2K$. Thus,$\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2K)}} = \frac{h}{\sqrt{16m_pK}} = \frac{h}{4\sqrt{m_pK}}$.Comparing the values:$\lambda_e = \sqrt{230} \cdot \frac{h}{\sqrt{2m_pK}} \approx 15.16 \cdot \frac{h}{\sqrt{2m_pK}}$$\lambda_p = \frac{h}{\sqrt{2m_pK}}$$\lambda_\alpha = \frac{1}{2\sqrt{2}} \cdot \frac{h}{\sqrt{2m_pK}} \approx 0.35 \cdot \frac{h}{\sqrt{2m_pK}}$Therefore,$\lambda_\alpha < \lambda_p < \lambda_e$.

The kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:

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