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(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.For relativistic electrons, the energy $E$ is related to momentum $p$ by $

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.For relativistic electrons, the energy $E$ is related to momentum $p$ by $E^2 = p^2c^2 + m_0^2c^4$. However, for high energy, $E \approx pc = \frac{hc}{\lambda}$.Given $\lambda = 7.5 	imes 10^{-12} \ m$.$E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{7.5 	imes 10^{-12}} \ J$.$E = \frac{19.89 	imes 10^{-26}}{7.5 	imes 10^{-12}} \ J = 2.652 	imes 10^{-14} \ J$.Converting to $eV$: $E = \frac{2.652 	imes 10^{-14}}{1.6 	imes 10^{-19}} \ eV \approx 1.65 	imes 10^5 \ eV = 165 \ keV$.Considering the non-relativistic approximation provided in the prompt's logic $(V \approx 25 \ kV)$, the question implies the use of the de Broglie relation $\lambda = \frac{h}{\sqrt{2mE}}$.$E = \frac{h^2}{2m\lambda^2} = \frac{(6.63 	imes 10^{-34})^2}{2 	imes 9.11 	imes 10^{-31} 	imes (7.5 	imes 10^{-12})^2} \ J$.$E \approx 4.3 	imes 10^{-15} \ J \approx 26.8 \ keV$.Thus, the closest option is $25 \ keV$.

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5 \times 10^{-12} \ m$, the minimum electron energy required is close to .............. $keV$.

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