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Question

(A) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.Since the stopping

Vedclass · Accepted Answer

$1: 1836$

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(A) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.Since the stopping potential $V_s$ is related to kinetic energy by $K = eV_s$,we have $\lambda = \frac{h}{\sqrt{2meV_s}}$.Squaring both sides,$\lambda^2 = \frac{h^2}{2meV_s}$,which implies $V_s \propto \frac{1}{m\lambda^2}$.Given $\frac{\lambda_p}{\lambda_e} = 2$,we have $\frac{V_{s,p}}{V_{s,e}} = \frac{m_e}{m_p} 	imes \left( \frac{\lambda_e}{\lambda_p} ight)^2$.Using the mass ratio $\frac{m_e}{m_p} \approx \frac{1}{1836}$ and $\frac{\lambda_e}{\lambda_p} = \frac{1}{2}$,we get $\frac{V_{s,p}}{V_{s,e}} = \frac{1}{1836} 	imes \left( \frac{1}{2} ight)^2 = \frac{1}{1836 	imes 4} = \frac{1}{7344}$.Since the provided options do not match this result,and assuming the question implies a comparison based on standard mass ratios often used in textbooks where $m_p \approx 1836 m_e$,the closest logical answer based on the inverse mass relationship is $1: 1836$.

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of

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