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In millikan's oil drop experiment the charge of three oil drops are 2,0.04 and 0.8 respectively. what could be the probable charge of electron and the number of electron attached to drop?
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In millikan's oil drop experiment the charge of three oil drops are 2,...
**Probable Charge of Electron:**

To determine the probable charge of an electron, we need to analyze the charges of the oil drops in Millikan's oil drop experiment. The basic idea behind this experiment is to measure the charge on individual oil droplets and compare it to the charge of an electron.

The charge on an oil drop is given by the equation:

q = n * e

Where q is the charge on the oil drop, n is the number of electrons attached to the drop, and e is the charge of an electron.

To find the charge of an electron, we can rearrange the equation as:

e = q / n

Now, let's use the given charges of the oil drops to calculate the probable charge of an electron.

**Charge of Oil Drop 1:**
q1 = 2
n1 = ?

Since the charge on the oil drop is given as 2, we need to find the number of electrons attached to the drop. Let's assume n1 is the number of electrons attached to drop 1.

Therefore, e = q1 / n1

We can rearrange this equation to find n1:

n1 = q1 / e

Substituting the values, we get:

n1 = 2 / e

**Charge of Oil Drop 2:**
q2 = 0.04
n2 = ?

Using the same approach as above, we can find n2:

n2 = q2 / e

Substituting the values, we get:

n2 = 0.04 / e

**Charge of Oil Drop 3:**
q3 = 0.8
n3 = ?

Using the same approach as above, we can find n3:

n3 = q3 / e

Substituting the values, we get:

n3 = 0.8 / e

**Comparing the Results:**

Now, let's compare the values of n1, n2, and n3 to find a common value for the number of electrons attached to all three drops. If we find a common value, it will give us the probable number of electrons attached to each oil drop and the probable charge of an electron.

By comparing n1, n2, and n3, we can see that the ratio of n1 to n2 is 50:1, and the ratio of n1 to n3 is 5:1. This implies that the number of electrons attached to oil drop 1 is 50 times the number of electrons attached to oil drop 2 and 5 times the number of electrons attached to oil drop 3.

Therefore, the most probable number of electrons attached to each oil drop is 50 for oil drop 1 and 10 for oil drop 3.

Now, we can calculate the charge of an electron by substituting the values of n1 and n3 into the equation:

e = q1 / n1 = 2 / 50 = 0.04

e = q3 / n3 = 0.8 / 10 = 0.08

By comparing these values, we can conclude that the probable charge of an electron is 0.04 or 0.08.
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In millikan's oil drop experiment the charge of three oil drops are 2,0.04 and 0.8 respectively. what could be the probable charge of electron and the number of electron attached to drop?
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