Let E_e and E_p represent the kinetic energy | vedclass

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(B) The de-Broglie wavelength of an electron is given by $\lambda_e = h / p_e = h / (m_e v)$.Given $v = c / 100$,so $\lambda_e = h / (m_e c / 100) = 1

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$10^2$

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(B) The de-Broglie wavelength of an electron is given by $\lambda_e = h / p_e = h / (m_e v)$.Given $v = c / 100$,so $\lambda_e = h / (m_e c / 100) = 100h / (m_e c)$.The kinetic energy of the electron is $E_e = (1/2) m_e v^2 = (1/2) m_e (c / 100)^2 = m_e c^2 / 20000$.The de-Broglie wavelength of a photon is $\lambda_p = h / p_p = hc / E_p$,which implies $E_p = hc / \lambda_p$.Given $\lambda_p = 2 \lambda_e$,we substitute $\lambda_e$:$\lambda_p = 2 	imes (100h / (m_e c)) = 200h / (m_e c)$.Now,calculate $E_p = hc / (200h / (m_e c)) = m_e c^2 / 200$.Finally,the ratio $E_p / E_e = (m_e c^2 / 200) / (m_e c^2 / 20000) = 20000 / 200 = 100 = 10^2$.

Let $E_e$ and $E_p$ represent the kinetic energy of an electron and a photon,respectively. If the de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron,then find the ratio $E_p / E_e$. (Given: speed of electron $v = c / 100$,where $c$ is the velocity of light).

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