Question

a proton and an alpha particle (a Helium atom stripped of its electrons) are accelerated through an electric potential difference of 20 MV. If the proton and alpha-particle start at rest, then what are their final speeds? answer is 62Mm/s (proton) and 44 Mm/s (alpha particle) .

Answer 1

Potential difference V = 20 x10 ⁶ volt

Initial speed u = 0

We know work done = (1/2) m[v ²- u ²]

                         Vq = (1/2)mv ²     SInce u = 0

From this velocity v = $\sqrt{}$ [2Vq/m]

For proton :   mass m = 1.67 x10 -27 kg

                     charge q = 1.6 x10 -19 C

Substitute values you get , v = $\sqrt{}$ [(2x20x10 ⁶ x1.6 x10 ^-19 ) /(1.67 x10 ^-27)]

                                             = 61.905 x10 ⁶ m/s

                                              = 62 x10 ⁶ m/s   (two significant figures)

                                             = 62 Mm/s

  

For alpha :   mass m = 4 x1.67 x10 ^-27 kg

                     charge q = 2 x1.6 x10 ^-19 C

Substitute values you get , v = $\sqrt{}$ [(2x20x10 ⁶ x2x1.6 x10 ^-19 ) /(4x1.67 x10 ^-27)]

                                             = 43.77 x10 ⁶ m/s

                                              = 44x10 ⁶ m/s   (two significant figures)

                                             = 44 Mm/s