The de Broglie wavelength of an electron movi | vedclass

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Question

(B) The de Broglie wavelength $\lambda$ of an electron is given by the formula:$\lambda = \frac{12.27}{\sqrt{K}} \; \mathring{A}$,where $K$ is the kin

Vedclass · Accepted Answer

$102 	imes 10^{-3} \;nm$

Vedclass · Answer

(B) The de Broglie wavelength $\lambda$ of an electron is given by the formula:$\lambda = \frac{12.27}{\sqrt{K}} \; \mathring{A}$,where $K$ is the kinetic energy in $eV$.Given $K = 144 \; eV$.Substituting the value of $K$ in the formula:$\lambda = \frac{12.27}{\sqrt{144}} \; \mathring{A}$$\lambda = \frac{12.27}{12} \; \mathring{A} = 1.0225 \; \mathring{A}$.Since $1 \; \mathring{A} = 0.1 \; nm$,we have:$\lambda = 1.0225 	imes 0.1 \; nm = 0.10225 \; nm$.Converting this to the form $102 	imes 10^{-x} \; nm$:$0.10225 \; nm = 102.25 	imes 10^{-3} \; nm$.Rounding to the nearest value,we get $102 	imes 10^{-3} \; nm$.

The de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

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