Question

An electron is accelerated to a speed of 4.23 × 10^6 ms –1 from rest by a potential difference. Calculate the potential difference

Answer 1

$e = 1.6*10^{-19}C$

$v = 4.23*10^6 \frac{m}{s}$

$m = 9.1*10^{-31}kg$

$\Delta \phi =$ ?

$A=\Delta E$

$e\Delta \phi = \large\frac{mv^2}{2} \to \Delta \phi = \large\frac{mv^2}{2e}=\frac{9.1*10^{-31}*(4.23)^2*10^{12}}{2*1.6*10^{-19}}=51V$