The kinetic energy of an electron and a proto | vedclass

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Question

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.Given that the kinetic energy $E = 10^{-32} \ J$ is t

Vedclass · Accepted Answer

$\lambda_p < \lambda_e$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$.Given that the kinetic energy $E = 10^{-32} \ J$ is the same for both the electron and the proton.Since $h$ and $E$ are constants,we have $\lambda \propto \frac{1}{\sqrt{m}}$.We know that the mass of a proton $(m_p)$ is greater than the mass of an electron $(m_e)$,i.e.,$m_p > m_e$.Therefore,$\sqrt{m_p} > \sqrt{m_e}$,which implies $\frac{1}{\sqrt{m_p}} Thus,$\lambda_p < \lambda_e$.

The kinetic energy of an electron and a proton is $10^{-32} \ J$. Then the relation between their de-Broglie wavelengths is

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