The distance between H+ and Cl- ions in HCl molecules is 1.38 mathringA. The potential due to this dipole at a distance of 10 mathringA on the axis of dipole is

Question

The distance between H+ and Cl- ions in HCl molecules is 1.38 mathringA. The potential due to this dipole at a distance of 10 mathringA on the axis of dipole is (A) 2.1 V (B) 1.8 V (C) 0.2 V (D) 1.2 V

Tardigrade Admin · Accepted Answer

Here, 2a=1.38×10-10 m, r=10 ×10-10 m charge, q=1.6×10-19 C As potential, V=(P/4πε0 r2)=(q(2a)/4πε0 r2) =(9×109×1.6×10-19×1.38×10-10/(10×10-10)2) =0.2 V

Q. The distance between $H^{+}$ and $C l^{-}$ ions in $H Cl$ molecules is $1.38 \overset{˚}{A}$ . The potential due to this dipole at a distance of $10 \overset{˚}{A}$ on the axis of dipole is

$2.1 V$

$1.8 V$

$0.2 V$

$1.2 V$

Q. The distance between H+ and Cl− ions in HCl molecules is 1.38A˚. The potential due to this dipole at a distance of 10A˚ on the axis of dipole is

2.1V

1.8V

0.2V

1.2V

Here, 2a=1.38×10−10m,r=10×10−10m charge, q=1.6×10−19C As potential, V=4πε0​r2P​=4πε0​r2q(2a)​ =(10×10−10)29×109×1.6×10−19×1.38×10−10​ =0.2V

Q. The distance between $H^{+}$ and $C l^{-}$ ions in $H Cl$ molecules is $1.38 \overset{˚}{A}$ . The potential due to this dipole at a distance of $10 \overset{˚}{A}$ on the axis of dipole is

$2.1 V$

$1.8 V$

$0.2 V$

$1.2 V$