Question

Suppose the charge of a proton and an electron differ slightly. One of them is $-e$, the other is $(e+\Delta e)$. If the resultant of electrostatic force and gravitational force between two hydrogen atoms placed at a distance $d$ (much greater than atomic size) apart is zero, then $\Delta e$ is of the order of [Given mass of hydrogen $m_h=1.67\times {10}^{-27}kg$]
[NEET- 2017]

• A. ${10}^{-37}C$
• B. ${10}^{-20}C$
• C. ${10}^{-47}C$
• D. ${10}^{-23}C$

Answer 1

The correct option is A ${10}^{-37}C$

Given,
$m_h=1.67\times {10}^{-27}kg$
And the distance between two hydrogen atoms is $d$.
Net charge on hydrogen atom $q=(e+\Delta e)-e=\Delta e$

Equating gravitational force and the electrostatic force (because the resultant of them is zero ), we get
$\frac{G{m}_h^2}{d^2}=\frac{k(\Delta e{)}^2}{d^2}$
$\frac{(6.67\times {10}^{-11})(1.67\times {10}^{-27}{)}^2}{d^2}=\frac{(9\times {10}^9)(\Delta e{)}^2}{d^2}$
$⟹\Delta e=2.06\times {10}^{-37}C$