The wavelength associated with an electron mo | vedclass

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Question

(B) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.Given: $h = 6.626 	imes 10^{-34} \, J \cdot s$,$m = 9.11 	imes 10^{-3

Vedclass · Accepted Answer

$7.25$

Vedclass · Answer

(B) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.Given: $h = 6.626 	imes 10^{-34} \, J \cdot s$,$m = 9.11 	imes 10^{-31} \, kg$,and $v = 10^8 \, cm/s = 10^6 \, m/s$.Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34} \, J \cdot s}{(9.11 	imes 10^{-31} \, kg) 	imes (10^6 \, m/s)}$$\lambda \approx 0.727 	imes 10^{-9} \, m = 7.27 	imes 10^{-10} \, m$.Since $1 \, \mathop {m{A}}\limits^{m{o}} = 10^{-10} \, m$,the wavelength is approximately $7.27 \, \mathop {m{A}}\limits^{m{o}}$,which is closest to $7.25 \, \mathop {m{A}}\limits^{m{o}}$.

The wavelength associated with an electron moving with a velocity of $10^8 \, cm/s$ is ........ $\mathop {\rm{A}}\limits^{\rm{o}} $.

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