Calculate the wavelength (in nanometer) assoc | vedclass

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(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$ Given: $h = 6.63 	imes 10^{-34} \ J \ s$ $m = 1.67 	imes

Vedclass · Accepted Answer

$0.40$

Vedclass · Answer

(A) The de Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$ Given: $h = 6.63 	imes 10^{-34} \ J \ s$ $m = 1.67 	imes 10^{-27} \ kg$ $v = 1.0 	imes 10^3 \ m \ s^{-1}$ Substituting the values: $\lambda = \frac{6.63 	imes 10^{-34}}{1.67 	imes 10^{-27} 	imes 1.0 	imes 10^3} \ m$ $\lambda = \frac{6.63}{1.67} 	imes 10^{-34 + 27 - 3} \ m$ $\lambda \approx 3.97 	imes 10^{-10} \ m$ Since $1 \ nm = 10^{-9} \ m$,we have: $\lambda \approx 0.397 	imes 10^{-9} \ m \approx 0.40 \ nm$

Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^3 \ m \ s^{-1}$. (Mass of proton $= 1.67 \times 10^{-27} \ kg$ and $h = 6.63 \times 10^{-34} \ J \ s$)

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