What is the experimental value of the de-Brog | vedclass

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(A) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula $\lambda = \frac{h}{p}$.

Vedclass · Accepted Answer

$\lambda = \frac{12.27}{\sqrt{V}} 	ext{ Å}$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula $\lambda = \frac{h}{p}$.Since the kinetic energy $K = eV = \frac{p^2}{2m}$, we have $p = \sqrt{2meV}$.Substituting this into the de-Broglie equation: $\lambda = \frac{h}{\sqrt{2meV}}$.Substituting the values of Planck's constant $h = 6.626 	imes 10^{-34} 	ext{ J s}$, mass of electron $m = 9.11 	imes 10^{-31} 	ext{ kg}$, and charge of electron $e = 1.602 	imes 10^{-19} 	ext{ C}$:$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 9.11 	imes 10^{-31} 	imes 1.602 	imes 10^{-19} 	imes V}}$.Calculating the constant factor, we get $\lambda \approx \frac{12.27 	imes 10^{-10}}{\sqrt{V}} 	ext{ m}$.Converting to $	ext{Å}$ $(1 	ext{ Å} = 10^{-10} 	ext{ m})$, we get $\lambda = \frac{12.27}{\sqrt{V}} 	ext{ Å}$.

What is the experimental value of the de-Broglie wavelength for an electron accelerated through a potential difference of $V$ volts?

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