The ratio of the de-Broglie wavelengths of a | vedclass

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Question

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2

Vedclass · Accepted Answer

$1: 43$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$.Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mK}}$.For a proton and an electron with the same kinetic energy $K$,the ratio of their wavelengths is $\frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_p K}}}{\frac{h}{\sqrt{2m_e K}}} = \sqrt{\frac{m_e}{m_p}}$.Given $m_p = 1849 	imes m_e$,we have $\frac{m_e}{m_p} = \frac{1}{1849}$.Therefore,$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{1}{1849}} = \frac{1}{43}$.Thus,the ratio is $1: 43$.

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)

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