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(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since $\lambda$ is the same for both the proton

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$1: 1836$

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(A) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since $\lambda$ is the same for both the proton and the electron,their momenta $p$ must also be equal.The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.Since $p$ is constant for both particles,we have $K \propto \frac{1}{m}$.Therefore,the ratio of the kinetic energy of the proton $(K_p)$ to the kinetic energy of the electron $(K_e)$ is:$\frac{K_p}{K_e} = \frac{m_e}{m_p}$.Given that $m_p = 1836 \ m_e$,we substitute this into the ratio:$\frac{K_p}{K_e} = \frac{m_e}{1836 \ m_e} = \frac{1}{1836}$.Thus,the ratio is $1: 1836$.

$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)

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$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)