Question

A proton is Accelerated with potential diffrence of 500 volt .Calculate the velocity of proton .​

Answer 1

Answer

here is the answer hope it's helpful

Answer 2

Given,

Proton accelerated with a potential difference(c) of 500 V.

To find,

the velocity of the proton which is accelerated (v)

Solution,

Now in this situation, we are assuming that

Mass of the proton (m): 1.66 × 10⁻²⁴ kg,

and charge of the proton(q) = 1.6 × 10⁻¹⁹ C.

We know the formula by which we can calculate the velocity,

$\frac{1}{2}$m$v^{2}$ = qc

⇒ $\frac{1}{2}$ × 1.66 × 10⁻²⁴ × $v^{2}$ = 500 × 1.6 × 10⁻¹⁹  

⇒ $v^{2}$ = 2 × 500 × 1.6 × 10⁻¹⁹ × 10²⁴ × $\frac{1}{1.66}$

⇒ $v^{2}$ = 1.6 × 10⁸ × $\frac{1}{1.66}$

⇒ $v^{2}$ = 0.9639 × 10⁸

⇒ v = $\sqrt{0.9639}$ × 10⁴

⇒ v = 0.9818 × 10⁴ m s⁻¹

Hence, the velocity of the given proton is 0.9818 × 10⁴ m s⁻¹.