Two protons are kept at a separation of 40 mathringA. Fn is the nuclear force and Fe, is the electrostatic force between them. Then

Question

Two protons are kept at a separation of 40 mathringA. Fn is the nuclear force and Fe, is the electrostatic force between them. Then (A) Fn >> Fe (B) Fn = Fe (C) Fn << Fe (D) Fn ≈ Fe

Tardigrade Admin · Accepted Answer

Fn is stronger than Fe . Fn operates at very short range inside the nucleus as little as 10-15 m As in the given case two protons are kept at a separation of 40 mathringA Fn << Fe

Q. Two protons are kept at a separation of 40 $\overset{˚}{A}$ . $F_{n}$ is the nuclear force and $F_{e}$ , is the electrostatic force between them. Then

$F_{n} >> F_{e}$

$F_{n} = F_{e}$

$F_{n} << F_{e}$

$F_{n} \approx F_{e}$

$F_{n}$ is stronger than $F_{e} . F_{n}$ operates at very short range inside the nucleus as little as $1 0^{- 15} m$ As in the given case two protons are kept at a separation of $40 \overset{˚}{A} F_{n} << F_{e}$

Q. Two protons are kept at a separation of 40 A˚. Fn​ is the nuclear force and Fe​, is the electrostatic force between them. Then

Fn​>>Fe​

Fn​=Fe​

Fn​<<Fe​

Fn​≈Fe​