KVPY 2020-21 SB SX Exam Previous Papers

Question : 23 of 120

Marks: +1, -0

A proton and an antiproton come close to each other in vacuum such that the distance between them is 10 cm. Consider the potential energy to be zero at infinity. The velocity at this distance will be :

1.17 m/s
2.3 m/s
3.0 m/s
23 m/s

Validate

Solution:

Applying mechanical energy conservation

{K}_{{i}}+{U}_{{i}}={K}_{{f}}+{U}_{{f}}

0+0=({1}/{2} {mv}^{2}+{1}/{2} {mv}^{2})+{{k}({q}_{1})({q}_{2})}/{{r}}

|{q}_{1}|=| {q}_{2} \|={e},

{q}_{1}=+{e}

{q}_{2}=-{e}

and r = 0.1 m

\∴{mv}^{2}={{ke}^{2}}/{{r}}

\∴{v}^{2}={{ke}^{2}}/{{mr}}={9 \×10^{9} \×(1.6 \×10^{-19})^{2}}/{(1.67 \×10^{-27})(0.1)}

{v}=1.17 {m} \/ {s}

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