Observe the following graph for the de-Brogli | vedclass

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Question

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Rearranging,we get $\lambda = \left(\frac{h}{\sqrt{2mq}}\right) V^{-1/2}$.

Vedclass · Accepted Answer

$45 	imes 10^{-49} \ kg$

Vedclass · Answer

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.Rearranging,we get $\lambda = \left(\frac{h}{\sqrt{2mq}}ight) V^{-1/2}$.The slope of the graph $\lambda$ vs $V^{-1/2}$ is $	an 	heta = \frac{h}{\sqrt{2mq}}$.Given $	an 	heta = 0.5$,$h = 6.0 	imes 10^{-34} \ J \cdot s$,and $q = 1.6 	imes 10^{-19} \ C$.$0.5 = \frac{6.0 	imes 10^{-34}}{\sqrt{2 	imes m 	imes 1.6 	imes 10^{-19}}}$.Squaring both sides: $0.25 = \frac{36 	imes 10^{-68}}{2 	imes m 	imes 1.6 	imes 10^{-19}}$.$m = \frac{36 	imes 10^{-68}}{0.25 	imes 3.2 	imes 10^{-19}} = \frac{36 	imes 10^{-68}}{0.8 	imes 10^{-19}} = 45 	imes 10^{-49} \ kg$.

Observe the following graph for the de-Broglie wavelength of a hypothetical charged particle $(q = 1.6 \times 10^{-19} \ C)$. Find the mass of the particle $(h = 6.0 \times 10^{-34} \ J \cdot s)$.

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