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Question: Two protons are placed \(10^{- 10}m\) apart. If they are repelled, what will be the kinetic energy o...

Two protons are placed 10−10m10^{- 10}m apart. If they are repelled, what will be the kinetic energy of each proton at very large distance

A

23×10−19J23 \times 10^{- 19}J

B

11.5×10−19J11.5 \times 10^{- 19}J

C

2.56×10−19J2.56 \times 10^{- 19}J

D

2.56×10−28J2.56 \times 10^{- 28}J

Answer

2.56×10−28J2.56 \times 10^{- 28}J

Explanation

Solution

Potential energy of the system when protons are separated by a distance of 10−1010^{- 10}m is

U=9×109×(1.6×10−19)210−10=23×10−19JU = \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{10^{- 10}} = 23 \times 10^{- 19}J

According to law of conservation of energy at very larger distance, this energy is equally distributed in both the protons as their kinetic energy hence K.E. of each proton will be 11.5×10−19J.11.5 \times 10^{- 19}J.