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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is t

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$\sqrt{1836} : 1$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass,and $E$ is the kinetic energy.Since the kinetic energy $E$ is the same for both the electron and the proton,we have:$\lambda_e = \frac{h}{\sqrt{2m_eE}}$ and $\lambda_p = \frac{h}{\sqrt{2m_pE}}$Taking the ratio of the wavelengths:$\frac{\lambda_e}{\lambda_p} = \frac{\sqrt{2m_pE}}{\sqrt{2m_eE}} = \sqrt{\frac{m_p}{m_e}}$Given that the mass of a proton $m_p \approx 1836 	imes m_e$,we substitute this value:$\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1836 	imes m_e}{m_e}} = \sqrt{1836}$Thus,the ratio $\lambda_e : \lambda_p = \sqrt{1836} : 1$.

If the kinetic energy of an electron and a proton are equal,find the ratio of their associated de Broglie wavelengths.

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