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(A) The kinetic energy $(K.E.)$ is given by $K.E. = \frac{p^2}{2m}$,where $p$ is momentum and $m$ is the mass of the electron.From this,$p = \sqrt{2mE

Vedclass · Accepted Answer

$7.7 	imes 10^{-8} \, cm$

Vedclass · Answer

(A) The kinetic energy $(K.E.)$ is given by $K.E. = \frac{p^2}{2m}$,where $p$ is momentum and $m$ is the mass of the electron.From this,$p = \sqrt{2mE}$.The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Given: $E = 2.5 \, eV = 2.5 	imes 1.6 	imes 10^{-12} \, erg$,$m = 9.1 	imes 10^{-28} \, g$,and $h = 6.626 	imes 10^{-27} \, erg \cdot s$.Substituting the values:$\lambda = \frac{6.626 	imes 10^{-27}}{\sqrt{2 	imes 9.1 	imes 10^{-28} 	imes 2.5 	imes 1.6 	imes 10^{-12}}}$$\lambda = \frac{6.626 	imes 10^{-27}}{\sqrt{72.8 	imes 10^{-40}}}$$\lambda = \frac{6.626 	imes 10^{-27}}{8.53 	imes 10^{-20}} \approx 7.7 	imes 10^{-8} \, cm$.

An electron moves in an electric field with a kinetic energy of $2.5 \, eV$. The de Broglie wavelength associated with it is .....

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