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(C) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.Given $K = 1.5 \ eV = 1.5

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$\lambda = 10^{-9} \ m$

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(C) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.Given $K = 1.5 \ eV = 1.5 	imes 1.6 	imes 10^{-19} \ J$.Using the formula $\lambda = \frac{12.27}{\sqrt{V}} \ \mathring{A}$,where $V$ is the accelerating potential in volts. Since $K = eV$,$V = 1.5 \ V$.$\lambda = \frac{12.27}{\sqrt{1.5}} \ \mathring{A} \approx \frac{12.27}{1.225} \ \mathring{A} \approx 10 \ \mathring{A}$.Converting to meters: $\lambda = 10 	imes 10^{-10} \ m = 10^{-9} \ m$.

An electron is accelerated from rest through a potential difference such that its kinetic energy is $1.5 \ eV$. The de-Broglie wavelength associated with this electron is:

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