The de Broglie wavelength of an electron of k | vedclass

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e K}}$.Given $K = 9 \ eV$ and $m_e c^2 = 0.5 \ MeV = 0.5 	imes

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$4 	imes 10^{-8} \ cm$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e K}}$.Given $K = 9 \ eV$ and $m_e c^2 = 0.5 \ MeV = 0.5 	imes 10^6 \ eV$.We know $m_e = \frac{0.5 	imes 10^6 \ eV}{c^2}$.Substituting this into the wavelength formula:$\lambda = \frac{h}{\sqrt{2 \left(\frac{0.5 	imes 10^6}{c^2}ight) K}} = \frac{hc}{\sqrt{2 	imes 0.5 	imes 10^6 	imes K}}$.Given $h = 4 	imes 10^{-15} \ eV \cdot s$ and $c = 3 	imes 10^{10} \ cm/s$,then $hc = (4 	imes 10^{-15}) 	imes (3 	imes 10^{10}) = 12 	imes 10^{-5} \ eV \cdot cm$.Now,$\lambda = \frac{12 	imes 10^{-5}}{\sqrt{2 	imes 0.5 	imes 10^6 	imes 9}} = \frac{12 	imes 10^{-5}}{\sqrt{9 	imes 10^6}} = \frac{12 	imes 10^{-5}}{3 	imes 10^3} = 4 	imes 10^{-8} \ cm$.

The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

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