Two protons are released at a distance of 0.1 nm from each other. What is their kinetic energy infinite distance apart ?

Question

Two protons are released at a distance of 0.1 nm from each other. What is their kinetic energy infinite distance apart ? (A) 23×1019J (B) 23×10-19J (C) 11.5×1019J (D) 11.5×10-19J

Tardigrade Admin · Accepted Answer

Initial P.E = (9 × 109 × ( 1.6 × 10-19)2/0.1 × 10-9) = 23 × 10-19 J It will be equally distributed between the two protons. So, the kinetic energy of each particle will be 11.5 × 10-19 J.

Q. Two protons are released at a distance of $0.1 nm$ from each other. What is their kinetic energy infinite distance apart ?

$23 \times 1 0^{19} J$

$23 \times 1 0^{- 19} J$

$11.5 \times 1 0^{19} J$

$11.5 \times 1 0^{- 19} J$

Initial P.E = $\frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{0.1 \times 1 0 ^{- 9}} = 23 \times 1 0^{- 19}$ J
It will be equally distributed between the two protons. So, the kinetic energy of each particle will be $11.5 \times 1 0^{- 19}$ J.

Q. Two protons are released at a distance of 0.1nm from each other. What is their kinetic energy infinite distance apart ?

23×1019J

23×10−19J

11.5×1019J

11.5×10−19J