Question

A proton (mp = 1.67x10-27 kg) is accelerated from rest through a potential difference of 750 volts. What is its final velocity?

Answer 1

3.79×10⁵ m/s

Step-by-step explanation

Given:

• The potential difference, ΔV = 750 V

,

• The mass of the proton, mp = 1.67*10⁻²⁷ kg

,

• The electric charge of a proton, e = 1.60*10⁻¹⁹ C

Unknown:

• The final velocity of the proton, v

The kinetic energy of the proton at the end of the motion is,

`KE=(1)/(2)m_pv^2`

This kinetic energy is obtained from the work done to move the proton,

`W=e\cdot\Delta V`

By conservation of energy,

`(1)/(2)m_pv^2=e\cdot\Delta V`

Solving for v,

`v=\sqrt[]{(2\cdot e\cdot\Delta V)/(m_p)}`

Replace with the known values and solve,

`v=\sqrt[]{(2\cdot1.60*10^(-19)C\cdot750V)/(1.67*10^(-27)kg)}\approx3.79*10^5m/s`

Hence, the final velocity of the proton is 3.79 × 10⁵ m/s.