Question

An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it

Answer 1

Answer:

V = 451.47 volts

Explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

a is acceleration, $a=\dfrac{v-u}{t}$ and u = 0

So,

$d=\dfrac{1}{2}(v-u)t$

$v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s$

Now applying the conservation of energy i.e.

$\dfrac{1}{2}mv^2=qV$

V is voltage

$V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V$

So, the voltage is 451.47 V.