a proton moves with a speed of 7.45 *10^5 directly towards a free proton initially at rest. find the distance of closest approach
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Let A proton of mass m moves with speed 7.45 × 10⁵ m/s directly towards another proton which is intially at rest .
we know, proton - proton repulsive force acts . both are repelling . Let r is the distance of closest approach .
Then, K.E energy of moving proton = electrostatic energy between protons
1/2mv² = Ke²/r
⇒ r = 2Ke²/mv² = 2 × 9 × 10⁹ × ( 1.6 × 10⁻¹⁹)²/1.675 × 10⁻²⁷ × (7.45 × 10⁵)we
[ Here mass of proton = 1.675 × 10⁻²⁷ Kg, e = 1.6 × 10⁻¹⁹C ]
∴ r = 5 × 10⁻¹³ m
Hence, distance of closest approach is 5 × 10⁻¹³ m
we know, proton - proton repulsive force acts . both are repelling . Let r is the distance of closest approach .
Then, K.E energy of moving proton = electrostatic energy between protons
1/2mv² = Ke²/r
⇒ r = 2Ke²/mv² = 2 × 9 × 10⁹ × ( 1.6 × 10⁻¹⁹)²/1.675 × 10⁻²⁷ × (7.45 × 10⁵)we
[ Here mass of proton = 1.675 × 10⁻²⁷ Kg, e = 1.6 × 10⁻¹⁹C ]
∴ r = 5 × 10⁻¹³ m
Hence, distance of closest approach is 5 × 10⁻¹³ m
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