Consider an electron (m = 9.1 times 10^{-31} | vedclass

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(B) For an electron confined between two rigid walls of length $L$,the particle acts like a standing wave. For the lowest energy state (ground state),

Vedclass · Accepted Answer

$6.0 	imes 10^{-20} \ J$

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(B) For an electron confined between two rigid walls of length $L$,the particle acts like a standing wave. For the lowest energy state (ground state),the length $L$ corresponds to half a wavelength: $L = \frac{\lambda}{2}$,so $\lambda = 2L$.Given $L = 1.0 	imes 10^{-9} \ m$,we have $\lambda = 2 	imes 10^{-9} \ m$.Using the de Broglie relation $p = \frac{h}{\lambda}$,the kinetic energy $E$ is given by $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9.1 	imes 10^{-31} \ kg$,and $\lambda = 2 	imes 10^{-9} \ m$:$E = \frac{(6.63 	imes 10^{-34})^2}{2 	imes (9.1 	imes 10^{-31}) 	imes (2 	imes 10^{-9})^2}$$E = \frac{43.96 	imes 10^{-68}}{18.2 	imes 10^{-31} 	imes 4 	imes 10^{-18}}$$E = \frac{43.96 	imes 10^{-68}}{72.8 	imes 10^{-49}} \approx 0.6038 	imes 10^{-19} \ J = 6.038 	imes 10^{-20} \ J$.Thus,the closest value is $6.0 	imes 10^{-20} \ J$.

Consider an electron $(m = 9.1 \times 10^{-31} \ kg)$ confined by electrical forces to move between two rigid walls separated by $1.0 \times 10^{-9} \ m$,which is about five atomic diameters. The quantized energy value for the lowest stationary state is

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