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The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer?
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The excess number of electrons that must be placed on each of two smal...
Given that,
Distance between two spheres , r = 3 cm 
We are recharging into SI unit , i.e. in metre.
Now , r = 3 x 0.01 = 0.3 m
Force of repulsion , F = 10^-19N

We need to find the number of elections by using the formula q=ne.
So,first let us use coulombs formula to find the value of q(charge).
F = k q^2/r^2
10^-19 = 9 x 10^9 x q^2 / 0.09
q^2 = 10^-19 x 0.09 / 9 x 10^9
q^2 = 10^-28 x 10^-2 
q= 10^-15 C.
Substitute it in q=ne to get n,
n = q/e
n = 10^-15/1.6x10^-19
n = 625.

Therefore , option (3) is correct.
Community Answer
The excess number of electrons that must be placed on each of two smal...
Calculation of Excess Number of Electrons

Given:
- Distance between the two spheres = 3cm
- Force of repulsion between the spheres = 10^-19 N

To find: Excess number of electrons that must be placed on each sphere

Solution:

We can use Coulomb's law to find the excess number of electrons on each sphere. Coulomb's law states that the force of repulsion between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Mathematically,

F = k * (q1 * q2) / r^2

where,
- F = Force of repulsion
- k = Coulomb's constant (9 x 10^9 Nm^2/C^2)
- q1 and q2 = Charges on the spheres
- r = Distance between the spheres

We can rearrange the formula to find the charge on one sphere.

q1 = F * r^2 / (k * q2)

Substituting the given values,

q1 = 10^-19 * (0.03)^2 / (9 x 10^9 * q2)

q1 = 10^-28 / (3 x 10^-9 * q2)

q1 = 10^-19 / q2

We know that the excess charge on each sphere is the same. Let's assume that the excess charge on each sphere is n times the charge of an electron (e).

q1 = n * e

Substituting this in the above equation,

n * e = 10^-19 / q2

n = (10^-19 / q2) / e

We know that the charge of an electron is 1.6 x 10^-19 C. Substituting this value,

n = (10^-19 / q2) / (1.6 x 10^-19)

n = 0.625 / q2

We can rearrange this equation to find q2.

q2 = 0.625 / n

Substituting the given answer option 2,

n = 225

q2 = 0.625 / 225

q2 = 2.78 x 10^-3 C

The excess charge on each sphere is

q1 = n * e = 225 * 1.6 x 10^-19

q1 = 3.6 x 10^-17 C

Therefore, the excess number of electrons on each sphere is

n = q1 / e = 3.6 x 10^-17 / 1.6 x 10^-19

n = 225

Hence, the correct answer is option 2 (225).
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Read the following text and answer the following questions on the basis of the same:Band theory of solid:Consider that the Si or Ge crystal contains N atoms. Electrons of each atom will have discrete energies in different orbits. The electron energy will be same if all the atoms are isolated, i.e., separated from each other by a large distance. However, in a crystal, the atoms are close to each other (2 Å to 3 Å) and therefore the electrons interact with each other and also with the neighbouring atomic cores. The overlap (or interaction) will be more felt by the electrons in the outermost orbit while the inner orbit or core electron energies may remain unaffected. Therefore, for understanding electron energies in Si or Ge crystal, we need to consider the changes in the energies of the electrons in the outermost orbit only. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of outer electrons in the orbit is 8 (2s + 6p electrons). So, out of the 4N electrons, 2N electrons are in the 2N s-states (orbital quantum number l = 0) and 2N electrons are in the available 6N p-states. Obviously, some p-electron states are empty. This is the case of well separated or isolated atoms.Q. The energy of electrons of atoms of a substance will be same if

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The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer?
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The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The excess number of electrons that must be placed on each of two small spheres spaced 3cm apart with force of repulsion b/w the spheres to be 10^-19 N is 1) 25 2) 225 3) 625 4) 1250 the correct answer is option 2. can you explain this answer?.
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