Two points A and B have potentials of 20 V a | vedclass

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(C) The potential difference through which the electron is accelerated is $\Delta V = V_B - V_A = 40 \ V - 20 \ V = 20 \ V$. The kinetic energy gained

Vedclass · Accepted Answer

$2.75 \ \mathring A$

Vedclass · Answer

(C) The potential difference through which the electron is accelerated is $\Delta V = V_B - V_A = 40 \ V - 20 \ V = 20 \ V$. The kinetic energy gained by the electron is $K = e \Delta V = 20 \ eV$. The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$. Substituting the values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$m = 9.11 	imes 10^{-31} \ kg$,and $K = 20 	imes 1.6 	imes 10^{-19} \ J$. $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.11 	imes 10^{-31} 	imes 20 	imes 1.6 	imes 10^{-19}}} \approx 2.75 	imes 10^{-10} \ m$. Therefore,$\lambda \approx 2.75 \ \mathring A$.

Two points $A$ and $B$ have potentials of $20 \ V$ and $40 \ V$ respectively. An electron is accelerated from rest between them. Find the de Broglie wavelength associated with the electron at point $B$.

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