Calculate the $(a)$ momentum,and $(b)$ de Broglie wavelength of the electrons accelerated through a potential difference of $56 \; V$.
(N/A) The momentum $p$ of an electron accelerated through a potential difference $V$ is given by the formula $p = \sqrt{2meV}$.
Substituting the values: $m = 9.11 \times 10^{-31} \; kg$,$e = 1.602 \times 10^{-19} \; C$,and $V = 56 \; V$.
$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times 56} \approx 4.04 \times 10^{-24} \; kg \cdot m/s$.
$(b)$ The de Broglie wavelength $\lambda$ is given by $\lambda = h/p$.
Using $h = 6.626 \times 10^{-34} \; J \cdot s$ and $p = 4.04 \times 10^{-24} \; kg \cdot m/s$:
$\lambda = (6.626 \times 10^{-34}) / (4.04 \times 10^{-24}) \approx 1.64 \times 10^{-10} \; m = 0.164 \; nm$.
Substituting the values: $m = 9.11 \times 10^{-31} \; kg$,$e = 1.602 \times 10^{-19} \; C$,and $V = 56 \; V$.
$p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times 56} \approx 4.04 \times 10^{-24} \; kg \cdot m/s$.
$(b)$ The de Broglie wavelength $\lambda$ is given by $\lambda = h/p$.
Using $h = 6.626 \times 10^{-34} \; J \cdot s$ and $p = 4.04 \times 10^{-24} \; kg \cdot m/s$:
$\lambda = (6.626 \times 10^{-34}) / (4.04 \times 10^{-24}) \approx 1.64 \times 10^{-10} \; m = 0.164 \; nm$.
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