The de-Broglie wavelength of an electron with | vedclass

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Question

(C) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_{e} K}}$, where $K$ is the kinetic energy

Vedclass · Accepted Answer

$62.5$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_{e} K}}$, where $K$ is the kinetic energy.Given: $h = 6.0 	imes 10^{-34} 	ext{ J s}$, $m_{e} = 9.0 	imes 10^{-31} 	ext{ kg}$, $K = 320 	ext{ eV} = 320 	imes 1.6 	imes 10^{-19} 	ext{ J}$.Substituting the values:$\lambda = \frac{6.0 	imes 10^{-34}}{\sqrt{2 	imes 9.0 	imes 10^{-31} 	imes 320 	imes 1.6 	imes 10^{-19}}}$$\lambda = \frac{6.0 	imes 10^{-34}}{\sqrt{18 	imes 10^{-31} 	imes 512 	imes 10^{-19}}}$$\lambda = \frac{6.0 	imes 10^{-34}}{\sqrt{9216 	imes 10^{-50}}}$$\lambda = \frac{6.0 	imes 10^{-34}}{96 	imes 10^{-25}}$$\lambda = 0.0625 	imes 10^{-9} 	ext{ m} = 62.5 	imes 10^{-12} 	ext{ m} = 62.5 	ext{ pm}$.

The de-Broglie wavelength of an electron with kinetic energy of $320 eV$ is (Take $h = 6.0 \times 10^{-34} \text{ SI unit}$, mass of electron $m_{e} = 9.0 \times 10^{-31} \text{ kg}$, charge of an electron $e = 1.6 \times 10^{-19} \text{ C}$). (in $pm$)

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