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Pie
3 years ago
8

In the Millikan oil drop experiment the measured charge of any single droplet was always a whole number multiple of -1.60 x 10-1

9 C. If a single droplet was found to have a total charge of -9.60×10-19 C then how many excess electrons are contained within the drop?
Physics
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

Number of electrons, n = 6

Explanation:

Total charge in a single droplet, q=-9.6\times 10^{-19}\ C

The measured charge of any single droplet, e=-1.6\times 10^{-19}\ C

Let n is the number of excess electrons are contained within the drop. According to the quantization of charge :

q=ne

n=\dfrac{q}{e}

n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}

n = 6

So, there are 6 electrons contained within the drop. Hence, this is the required solution.

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How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

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3 years ago
A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r
Hatshy [7]

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The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

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work = V.Q

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w= 12 . 50.10^-6
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Answer:

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