If the kinetic energy of an electron is 1.25 | vedclass

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(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mK.E.}}$Given:$h = 6.6 	imes 10^{-34} \ Js$$m = 9 	imes

Vedclass · Accepted Answer

$4.4 	imes 10^{-5} \ m$

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(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mK.E.}}$Given:$h = 6.6 	imes 10^{-34} \ Js$$m = 9 	imes 10^{-31} \ kg$$K.E. = 1.25 	imes 10^{-28} \ J$Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 1.25 	imes 10^{-28}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{22.5 	imes 10^{-59}}} = \frac{6.6 	imes 10^{-34}}{\sqrt{225 	imes 10^{-60}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{15 	imes 10^{-30}} = 0.44 	imes 10^{-4} \ m = 4.4 	imes 10^{-5} \ m$

If the kinetic energy of an electron is $1.25 \times 10^{-28} \ J$,then what will be the de-Broglie wavelength of the electron? (Mass of electron $= 9 \times 10^{-31} \ kg$,$h = 6.6 \times 10^{-34} \ Js$)

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