If the de Broglie wavelength of an electron i | vedclass

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(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.Rearranging for $

Vedclass · Accepted Answer

$4.55 	imes 10^{-25}$

Vedclass · Answer

(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy.Rearranging for $K$,we get $K = \frac{h^2}{2m\lambda^2}$.Given: $\lambda = 728.14 \ nm = 728.14 	imes 10^{-9} \ m$,$m = 9.1 	imes 10^{-31} \ kg$,and $h = 6.626 	imes 10^{-34} \ J \ s$.Substituting the values:$K = \frac{(6.626 	imes 10^{-34})^2}{2 	imes 9.1 	imes 10^{-31} 	imes (728.14 	imes 10^{-9})^2}$$K = \frac{43.903876 	imes 10^{-68}}{18.2 	imes 10^{-31} 	imes 530187.9 	imes 10^{-18}}$$K = \frac{43.903876 	imes 10^{-68}}{96.494 	imes 10^{-25}}$$K \approx 0.455 	imes 10^{-24} \ J = 4.55 	imes 10^{-25} \ J$.Thus,the correct option is $A$.

If the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

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