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Question: In an oil drop experiment, the following charges (in arbitrary units) were found on a series of oil ...

In an oil drop experiment, the following charges (in arbitrary units) were found on a series of oil droplets 2.30×10−152.30 \times {10^{ - 15}}, 6.90×10−156.90 \times {10^{ - 15}}, 1.38×10−141.38 \times {10^{ - 14}}, 5.75×10−155.75 \times {10^{ - 15}}, 3.45×10−153.45 \times {10^{ - 15}}, 1.96×10−141.96 \times {10^{ - 14}}. The magnitude of charge on the electron (in the same unit) is

Explanation

Solution

. Refer to the conclusion of the oil drop experiment given by R.A. Mulliken. The magnitude of charge on the electron is always an integral multiple of the smallest electrical charge. Find out the possible smallest number which when multiplied by a positive integer gives the charges given in our experimental data.

Complete step by step answer:
Oil drop experiment was given by R.A. Millikan to determine the charge on the electrons. A conclusion of his oil drop experiment was that the magnitude of charge on the electron is always an integral multiple of the smallest electrical charge. So, in the given data of electrical charges that is, 2.30×10−152.30 \times {10^{ - 15}}, 6.90×10−156.90 \times {10^{ - 15}}, 1.38×10−141.38 \times {10^{ - 14}}, 5.75×10−155.75 \times {10^{ - 15}}, 3.45×10−153.45 \times {10^{ - 15}}, 1.96×10−141.96 \times {10^{ - 14}}, the smallest charge is 2.30×10−152.30 \times {10^{ - 15}}. But you can find that all other given charges are not multiple of this smallest charge. So, we need to find the required smallest number. Let us divide 2.30×10−152.30 \times {10^{ - 15}} by 2, we get 1.15×10−151.15 \times {10^{ - 15}}.

Let us find out if all the given charges (in the same unit) are integral multiples of 1.15×10−151.15 \times {10^{ - 15}}.
2.30×10−152.30 \times {10^{ - 15}} is 2×(1.15×10−15)2 \times (1.15 \times {10^{ - 15}}). Thus, 2.30×10−152.30 \times {10^{ - 15}} is an integral multiple of 1.15×10−151.15 \times {10^{ - 15}}. 6.90×10−156.90 \times {10^{ - 15}} is six times the (1.15×10−15)(1.15 \times {10^{ - 15}}), 5.75×10−155.75 \times {10^{ - 15}} is five times the 1.15×10−151.15 \times {10^{ - 15}}and 3.45×10−153.45 \times {10^{ - 15}} is three times the charge of electron i.e., 1.15×10−151.15 \times {10^{ - 15}}. Hence, all the given charges (in the same unit) are integral multiples of 1.15×10−151.15 \times {10^{ - 15}}.

Thus, the required magnitude of charge on the electron (in the same unit) is 1.15×10−151.15 \times {10^{ - 15}}.

Note: Mullikan in his oil drop experiment found that the charge on the electron i.e., (ee) to be −1.6×10−19C - 1.6 \times {10^{ - 19}}C. Mullikan concluded that the magnitude of electrical charge, qq, on the droplets is always an integral multiple of the electrical charge, ee, that is, q=neq = ne where, n = 1, 2, 3... .