If the kinetic energy of an electron is 18.2 | vedclass

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Question

(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is t

Vedclass · Accepted Answer

$364$

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(C) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $K$ is the kinetic energy.Given: $h = 6.626 	imes 10^{-34} \ J \ s$,$m = 9.1 	imes 10^{-31} \ kg$,$K = 18.2 	imes 10^{-25} \ J$.Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 18.2 	imes 10^{-25}}}$$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{331.24 	imes 10^{-56}}}$$\lambda = \frac{6.626 	imes 10^{-34}}{18.2 	imes 10^{-28}}$$\lambda = 0.364 	imes 10^{-6} \ m = 364 	imes 10^{-9} \ m = 364 \ nm$.

If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

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