Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 162 V

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(a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 162 V.

m/s

(b) Calculate the speed of an electron that is accelerated through the same potential difference.
m/s
asked Jan 20, 2012 in Physics by Roar3215 ~Expert~ (1,298 points)
    

1 Answer

+2 votes
 
Best answer
To solve this problem use the law of conservation of energy:

Ki + Ui = Kf + Uf  (where K = kinetic energy and U = potential ,  i = initial, f = final)

now substitute the values accordingly:

0 + qV = (1/2)*m*v^2 + 0

recall that, q = 1.60*10^-19 C, V = 162 V, and m = 1.673*10^-27 kg

Now substitute in this values to above equation.

(1.60*10^-19 C)*(162 V)*(1J / 1V*C) = (1/2)*(1.673*10^-27 kg)*v^2

*note that the (1J / 1V*C) on the left does not change the equation, it only converts the units to J,

Now rearrange and solve for v:

v = squareroot of [(2*((1.60*10^-19 C)*(162 V)*(1J / 1V*C)) / (1.673*10^-27 kg))]

and hence v = 176,029 m/s

 

Part B, is solved in the same way, only this time the mass is changed to 9.109*10^-31 kg, which is the mass of one electron. the charge remains the same, because they the magnitude is equal for both proton and electron

and so,
v = squareroot of [(2*((1.60*10^-19 C)*(162 V)*(1J / 1V*C)) / (9.109*10^-31 kg))]

and you get v = 7,543,921 m/s
answered Jan 24, 2012 by pokemonmaster ~Expert~ (3,842 points)
selected Jan 24, 2012 by Roar3215

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