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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{P}$.Since the proton and electron have the same wavelength,$\lambda_p = \lambda_e$,it im

Vedclass · Accepted Answer

${K}_{p} < {K}_{e}$ and ${P}_{p} = {P}_{e}$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{P}$.Since the proton and electron have the same wavelength,$\lambda_p = \lambda_e$,it implies that their momenta are equal: ${P}_p = {P}_e$.Kinetic energy is related to momentum by the formula $K = \frac{P^2}{2m}$.For the proton: ${K}_p = \frac{{P}_p^2}{2{m}_p}$.For the electron: ${K}_e = \frac{{P}_e^2}{2{m}_e}$.Since ${P}_p = {P}_e$,the ratio of kinetic energies is $\frac{{K}_p}{{K}_e} = \frac{{m}_e}{{m}_p}$.Because the mass of a proton is much greater than the mass of an electron $({m}_p > {m}_e)$,it follows that ${K}_p < {K}_e$.

$A$ moving proton and electron have the same de Broglie wavelength. If ${K}$ and ${P}$ denote the kinetic energy and momentum respectively,then choose the correct option:

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