What is the de-Broglie wavelength associated | vedclass

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(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.Given values are $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m_{

Vedclass · Accepted Answer

$0.114$

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(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.Given values are $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m_{e} = 9.11 	imes 10^{-31} \ kg$,and $v = 6.4 	imes 10^{6} \ m/s$.Substituting these values into the formula:$\lambda = \frac{6.63 	imes 10^{-34}}{9.11 	imes 10^{-31} 	imes 6.4 	imes 10^{6}}$$\lambda = \frac{6.63 	imes 10^{-34}}{58.304 	imes 10^{-25}}$$\lambda \approx 0.1137 	imes 10^{-9} \ m$Since $1 \ nm = 10^{-9} \ m$,we get $\lambda \approx 0.114 \ nm$.

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

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