Question

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 2V0, what speed would it gain? A proton is accelerated from rest through a potential difference and gains a speed . If it were accelerated instead through a potential difference of , what speed would it gain? v02√ 4v0 8v0 2v0 SubmitReques

Answer 1

Answer:

The speed is $\sqrt{2}v_{0}$.

(a) is correct option.

Explanation:

Given that,

Potential difference $V= V_{0}$

Speed $v = v_{o}$

If it were accelerated instead

Potential difference $V'=2V_{0}$

We need to calculate the speed

Using formula of initial work done on proton

$W = q V$

We know that,

$\Delta W=\Delta K.E$

$q V=\dfrac{1}{2}mv^2$

Put the value into the formula

$q V_{0}=\dfrac{1}{2}mv_{0}^2$

$v_{0}^2=\dfrac{2qV_{0}}{m}$....(I)

If it were accelerated instead through a potential difference of $2 V_{0}$, then it would gain a speed will be given as :

Using an above formula,

$v_{0}'^2=\dfrac{2qV_{0}}{m}$

Put the value of $V_{0}$

$v_{0}'^2=\dfrac{2q\times2V_{0}}{m}$

$v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}$

$v_{0}'=\sqrt{2}v_{0}$

Hence, The speed is $\sqrt{2}v_{0}$.