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Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of
[given ma of hydrogen mh = 1.67×10−27kg]
  • a)
    10−23C
  • b)
    10−37C
  • c)
    10−47C
  • d)
    10−20C
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
Suppose the charge of a proton and an electron differ slightly. One o...
Explanation:

To find the value of Δe, we need to consider the electrostatic force and gravitational force between two hydrogen atoms placed at a distance d apart. Let's break down the problem into smaller steps:

Step 1: Electrostatic Force between Two Hydrogen Atoms:
The electrostatic force between two charged particles is given by Coulomb's Law:

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

Where:
F_e is the electrostatic force
k is the electrostatic constant (9 × 10^9 N m^2/C^2)
q1 and q2 are the charges of the particles (in this case, -e and e Δe)
r is the distance between the particles (d)

For hydrogen atoms, the charge of a proton is +e and the charge of an electron is -e. So, the electrostatic force between two hydrogen atoms is:

F_e = (k * (+e) * (-e Δe)) / d^2
= -k * e^2 Δe / d^2

Step 2: Gravitational Force between Two Hydrogen Atoms:
The gravitational force between two objects is given by Newton's Law of Universal Gravitation:

F_g = (G * m1 * m2) / r^2

Where:
F_g is the gravitational force
G is the gravitational constant (6.67 × 10^-11 N m^2/kg^2)
m1 and m2 are the masses of the objects (in this case, mass of hydrogen atom mh)
r is the distance between the objects (d)

For two hydrogen atoms, the gravitational force is:

F_g = (G * mh * mh) / d^2
= G * mh^2 / d^2

Step 3: Equating the Forces:
Given that the net force between the two hydrogen atoms is zero, we can equate the electrostatic force and the gravitational force:

-k * e^2 Δe / d^2 = G * mh^2 / d^2

Simplifying the equation:

Δe = (G * mh^2 * d^2) / (-k * e^2)

Plugging in the given values:

Δe = (6.67 × 10^-11 N m^2/kg^2) * (1.67 × 10^-27 kg)^2 * (d^2) / (-9 × 10^9 N m^2/C^2) * (1.6 × 10^-19 C)^2

Simplifying further:

Δe = (1.11 × 10^-96 kg^2 m^2) / (2.304 × 10^-28 C^2)
= 4.81 × 10^-69 C

The order of magnitude of Δe is 10^-69 C, which is closest to option B (10^-37 C).
Free Test
Community Answer
Suppose the charge of a proton and an electron differ slightly. One o...
A hydrogen atom consists of an electron and a proton.
∴ Charge on one hydrogen atom
= qe + qp = −e + (e + Δe) = Δe
Since a hydrogen atom carry a net charge Δe
∴ Electrostatic force
will act between two hydrogen atoms.
The gravitational force between two hydrogen atoms is given as
Since, the net force on the system is zero,Fe = Fg
Using eqns. (i) and (ii), we get
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Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of[given ma of hydrogen mh = 1.67×10−27kg]a)10−23Cb)10−37Cc)10−47Cd)10−20CCorrect answer is option 'B'. Can you explain this answer?
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Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of[given ma of hydrogen mh = 1.67×10−27kg]a)10−23Cb)10−37Cc)10−47Cd)10−20CCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of[given ma of hydrogen mh = 1.67×10−27kg]a)10−23Cb)10−37Cc)10−47Cd)10−20CCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e+Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of[given ma of hydrogen mh = 1.67×10−27kg]a)10−23Cb)10−37Cc)10−47Cd)10−20CCorrect answer is option 'B'. Can you explain this answer?.
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