Problem: A proton has an initial speed of 3.9 × 105 m/s.What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 3?_____ kV

FREE Expert Solution

Electrical potential energy:

$\boxed{\mathbf{U}\mathbf{=}\mathbf{q}\mathbf{V}}$

Kinetic energy:

$\boxed{\mathbf{K}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}$

U₀ = qV₀

U_f = qV_f 

KE₀ = (1/2)mv₀² 

KE_f = [(1/2)mv₀²]/3 = (1/6)mv₀²