A particle having electric charge 3 times 10^ | vedclass

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Question

(C) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqr

Vedclass · Accepted Answer

$10$

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(C) The de Broglie wavelength $\lambda$ of a particle accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mqV}}$.Given: $h = 6.6 	imes 10^{-34} 	ext{ J} \cdot 	ext{s}$, $q = 3 	imes 10^{-19} 	ext{ C}$, $m = 6 	imes 10^{-27} 	ext{ kg}$, and $V = 1.21 	ext{ V}$.Substituting the values into the formula:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes (6 	imes 10^{-27}) 	imes (3 	imes 10^{-19}) 	imes 1.21}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{36 	imes 10^{-46} 	imes 1.21}}$$\lambda = \frac{6.6 	imes 10^{-34}}{6 	imes 10^{-23} 	imes 1.1}$$\lambda = \frac{6.6 	imes 10^{-34}}{6.6 	imes 10^{-23}}$$\lambda = 10^{-11} 	ext{ m} = 10 	imes 10^{-12} 	ext{ m}$.Comparing this with $\alpha 	imes 10^{-12} 	ext{ m}$, we get $\alpha = 10$.

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