A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V.What speed does the proton gain? (e = 1.60 × 10^-19C,m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

We can use the formula for the kinetic energy gained by a charged particle moving through an electric potential difference:
∆K = q∆V
where ∆K is the change in kinetic energy, q is the charge of the particle, and ∆V is the potential difference.
Plugging in the values, we have:
∆K = (1.60 × 10^-19 C)(500 V) = 8.0 × 10^-17 J
We can equate this to the kinetic energy of the proton:
∆K = (1/2)mv^2
where m is the mass of the proton and v is its final velocity.
Solving for v, we get:
v = √(2∆K/m) = √(2(8.0 × 10^-17 J)/(1.67 × 10^-27 kg)) ≈ 3.1 × 10^5 m/s
Therefore, the proton gains a final speed of 3.1 × 10^5 m/s. The answer is choice B).

A Proton That Is Initially at Rest Is Accelerated Through