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valentina_108 [34]
3 years ago
14

As in the video, we apply a charge + to the half-shell that carries the electroscope. This time, we also apply a charge – to the

other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?
A) It deflects more than it did at the end of the video.
B) It does not deflect at all.
C) It deflects the same amount as at end of the video.
D) It deflects less than it did at the end of the video.
Physics
1 answer:
horrorfan [7]3 years ago
8 0

Note: Although the video is not provided in this question, it is not needed to answer the question.

Answer:

B) It does not deflect at all

Explanation:

Since both half shells contain opposite charges, the two shells become electrically neutral when they are brought together and the electroscope discharges. On separating the two half shells again, the needle does not deflect because the half shells have now lost their charges to become neutral.

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Answer:

The edge length is 0.4036 nm

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As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

V_{c} = a^{3}  = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

Therefore, the length of the unit cell is given as:

a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}            (1)

Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

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C_{Ag} = 79 %

A_{Ag} = 107

C_{Pd} = 21%

A_{Pd} = 106

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A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

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