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(D) The De Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is mass,and $E

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$E_{\alpha} < E_d < E_p$

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(D) The De Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is mass,and $E$ is kinetic energy. Since $\lambda$ is the same for all,we have $E = \frac{h^2}{2m\lambda^2}$. This implies $E \propto \frac{1}{m}$. The masses of the particles are: $m_p = m$,$m_d = 2m$,and $m_{\alpha} = 4m$. Therefore,the kinetic energies are $E_p = \frac{k}{m}$,$E_d = \frac{k}{2m}$,and $E_{\alpha} = \frac{k}{4m}$ (where $k = \frac{h^2}{2\lambda^2}$). Comparing these values,we get $E_{\alpha} < E_d < E_p$.

If a moving proton,deuteron,and alpha particle have the same De Broglie wavelength,then what is the correct order of their kinetic energy?

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