A scientist repeats the millikan oil drop experiment in a different galaxy and the charge on the drops is measured in a unit called the electrino (el). The scientist obtains data for four drops.

Respuesta :

The el unit obtained is 0.46 x 10 ^ - 14 el and is usually obtained by a cost that elevated via way of means of special integers offers us the values of the expenses given.

For this we seize the smallest fee, divide it via way of means of different integers (for the reason that elemental fee have to follow this condition) and multiply it via way of means of those cited at the start to try and gain the alternative expenses. We have to count on consequences that have been in no way measured additionally (which "may be").

  • For now shall we forget about all the
  • 10 ^ - 14 and the units.
  • We begin then with 1.84. Maybe that is the basic fee, and we might most effective want to multiply via way of means of different integers to gain the relaxation of the expenses. Lets see:
  • 2(1.84)=three.sixty eight, we can not get three.22, we will attempt with the following possibility,:
  • 2(1.84/2) = 1.84 , obviously,
  • 3(1.84/2) = 2.76 may be,
  • 4(1.84/2) = 6.68 once more we can not get 3.22, we can attempt with the following possibility,
  • 2(1.84/three) = 1.23 may be,
  • 3(1.84/three) = 1.23 , obviously,
  • 4(1.84/three) = 2.45 may be,
  • 5(1.84/three) = 3.07 may be,
  • 6(1.84/three) = 3.68 , once more we can not get three.22, we can attempt with the following possibility,
  • 2(1.84/4) = 0.92 , may be,
  • 3(1.84/4) = 1.38 may be,
  • 4(1.84/4) = 1.84 , obviously,
  • 5(1.84/4) = 2.3 may be,
  • 6(1.84/4) = 2.76 , may be,
  • 7(1.84/4) = three.22 we received any other cost! Two to go,
  • 8(1.84/4) = 3.68, may be,
  • 9(1.84/4) = 4.14 we received any other cost! One to go,
  • 10(1.84/4) = 4.6 may be,
  • 1(1.84/4) = 5.06 may be,
  • 1(1.84/4) = 5.2 done!
  • Since (1.84/4) = 0.46 , the basic fee (fee of the electron) may be 0.46x 10 ^ - 14 el.

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