The de-Broglie wavelength of a proton (charge | vedclass

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Question

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.S

Vedclass · Accepted Answer

$0.9 	imes 10^{-12} \ m$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.Substituting the values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 1.67 	imes 10^{-27} \ kg$,$q = 1.6 	imes 10^{-19} \ C$,and $V = 1000 \ V$.$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 1.67 	imes 10^{-27} 	imes 1.6 	imes 10^{-19} 	imes 1000}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{5.344 	imes 10^{-43}}} = \frac{6.63 	imes 10^{-34}}{7.31 	imes 10^{-22}} \approx 0.9 	imes 10^{-12} \ m$.

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{-19} \ C$,mass $= 1.67 \times 10^{-27} \ kg$) accelerated through a potential difference of $1 \ kV$ is:

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