A proton has an initial speed of 3.5×10 5 m/s. a). What potential difference is

Question

A proton has an initial speed of 3.5×105 m/s.

a). What potential difference is required to bring the proton to rest?

b). What potential difference is required to reduce the initial speed of the proton by a factor of 2?

c). What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2? A proton has an initial speed of 3.5×105 m/s.

a). What potential difference is required to bring the proton to rest?

b). What potential difference is required to reduce the initial speed of the proton by a factor of 2?

c). What potential difference is required to reduce the initial kinetic energy of the proton by a factor of 2?

Explanation / Answer

Initial Kinetic Energy K.Ei= 0.5*m*u^2 = 0.5*1.672*10^(-27)*(3.5×10^5)^2 = 1.024*10^(-16) J a) Final Kinetic Energy K.Ef= 0 qV + 0= K.E V = 1.024*10^(-16)/(1.602*10^(-19)) = 639.26 V b) Final Kinetic Energy K.Ef = 0.5*1.672*10^(-27)*((3.5/2)×10^5)^2 = 0.256*10^(-16) J qV + K.Ef = K.Ei V = (1.024-0.256)*10^(-16)/(1.602*10^(-19)) = 479.445 V c)Final Kinetic Energy K.Ef = (0.5*1.672*10^(-27)*(3.5×10^5)^2 )/2= 0.512*10^(-16) J qV + K.Ef = K.Ei V = (1.024-0.512)*10^(-16)/(1.602*10^(-19)) = 319.63 V

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