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Atoms and Nuclei question

2022 · Q176
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Atoms and Nuclei question

2022 · Q176

NEETPhysicsAtoms and NucleiMCQ+4 / −1

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is 2.4 ×\times× 1039. The ratio of the proportionality constant, K=14πε0K = {1 \over {4\pi {\varepsilon _0}}}K=4πε0​1​ to the gravitational constant G is nearly (Given that the charge of the proton and electron each = 1.6 ×\times× 10−-−19 C, the mass of the electron = 9.11 ×\times× 10−-−31 kg, the mass of the proton = 1.67 ×\times× 10−-−27 kg) :

  1. A
    10
  2. B
    1020
  3. C
    1030
  4. D
    1040
View written solutionFree

Correct answer: B

Ratio of magnitude of Coulomb's electrostatic force to the gravitational force

$${{{F_E}} \over {{F_G}}} = {{\left( {{{K{q_1}{q_2}} \over {{r^2}}}} \right)} \over {\left( {{{G{m_1}{m_2}} \over {{r^2}}}} \right)}} = {{K{q_1}{q_2}} \over {G{m_1}{m_2}}}$$

$$ \Rightarrow 2.4 \times {10^{39}} = {K \over G} \times {{1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}} \over {9.11 \times {{10}^{ - 31}} \times 1.67 \times {{10}^{ - 27}}}}$$

$$ \Rightarrow 2.4 \times {10^{39}} = {K \over G} \times {{2.56} \over {15.21}} \times {10^{20}} \Rightarrow {K \over G} = 14.26 \times {10^{19}}$$

$$ \Rightarrow {K \over G} = 1.426 \times {10^{20}}$$ $\Rightarrow$ Ratio $\approx$ 1020

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